Question 15 Marks
Prove the following by using the principle of mathematical induction for all $n \in N:$
$n(n + 1)(n + 5)$ is a multiple of $3.$
AnswerLet $P(n)=n(n+1)(n+5)$ is a multiple of 3 . For $n=1, P(1)=1(1+1)(1+5)$ is a multiple of $3=12$ is a multiple of 3 $\therefore P (1)$ is true.
Let P(n) be true for $n = k, P(k) = k(k + 1)(k + 5)$ is a multiple of $3.$
$\Rightarrow\ \text{k}(\text{k+1})(\text{k+5})=3\lambda$
$\Rightarrow\ \text{k}^3+6\text{k}^2+5\text{k}=3\lambda$
$\Rightarrow\ \text{k}^3=3\lambda-6\text{k}^2-5\text{k}\ ....(\text{i})$
For $n = k + 1, P(k + 1) = (k + 1)(k + 2)(k + 6) $is a multiple of $3$
Now, $(k + 1)(k + 2)(k + 6) = k^3 + 9k^2 + 20k + 12$
$=3\lambda-6\text{k}^2-5\text{k}+9\text{k}^2+20\text{k}+12\ \ [\text{Using (i)}]$
$=3\lambda+3\text{k}^2+15\text{k}+12$
$=3(\lambda+\text{k}^2+5\text{k}+4)$
$= (k + 1)(k + 2)(k + 6)$ is a multiple of 3 $\therefore$ P(k + 1) is true. Therefore, P(k) is true.
$\Rightarrow P(k + 1)$ is true.
Hence by Principle of Mathematical Induction, P(n) is true for all $\text{n}\in\text{N.}$
View full question & answer→Question 25 Marks
Prove the following by using the principle of mathematical induction for all $n \in N:$
$41^n – 14^n $ is a multiple of $27.$
AnswerLet $P(n)=41^n-14^n$ is a multiple of 27 . For $n=1, P(1)=41^1-14^1$ is a multiple of $27=27$ is a multiple of $27 \therefore P(1)$ is true. Let $P(n)$ be true for $n=k, P(k)=41^k-14^k$ is a multiple of 27 .
$41^{\text{k}}-14^{\text{k}}=27\lambda\ ....(\text{i})$
For n = k + 1,
$\text{P(k+1)}=41^{\text{k+1}}-14^{\text{k+1}}$ is a multiple of 27.Now, $41^{\text{k+1}}-14^{\text{k+1}}$
$=41^{\text{k+1}}-41^{\text{k}}.14+41^{\text{k}.14}-14^{\text{k+1}}$
$=41^{\text{k}}(41-14)+14(41^{\text{k}}-14^\text{k})$
$=41^{\text{k}}\times27+14\times27\lambda\ \ [\text{From (i)}]$
$=27(41^{\text{k}}+14\lambda)$
$\therefore\ 41^{\text{k+1}}-14^{\text{k+1}}$ is a multiple of 27.
$\therefore\ \text{P(k+1)}$ is true. Therefore, P(k) is true. ⇒ P(k + 1) is true.Hence by Principle of Mathematical Induction, P(n) is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 35 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{n}}=1-\frac{1}{2^{\text{n}}}.$
AnswerLet the given statement be P(n), i.e., $\text{P(n)}:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{n}}=1-\frac{1}{2^{\text{n}}}$ For n = 1, we have$\text{P(1)}:\frac{1}{2}=1-\frac{1}{2^1}=\frac{1}{2},$
which is true. Let P(k) be true for some positive integer k, i.e.,$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{k}}=1-\frac{1}{2^{\text{k}}}\ \ ....(\text{i})$
We shall now prove that P(k + 1) is true. Consider, $\Big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{2^{\text{k}}}\Big)+\frac{1}{2^{\text{k+1}}}$ $=\Big(1-\frac{1}{2^{\text{k}}}\Big)+\frac{1}{2^{\text{k+1}}}\ \ [\text{Using (i)}]$ $=1-\frac{1}{2^{\text{k}}}+\frac{1}{2.2^{\text{k}}}$ $=1-\frac{1}{2^{\text{k}}}\Big(1-\frac{1}{2}\Big)$ $=1-\frac{1}{2^{\text{k}}}\Big(\frac{1}{2}\Big)$ $=1-\frac{1}{2^{\text{k+1}}}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 45 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:
$1+2+3+...+\text{n}<\frac{1}{8}(2\text{n}+1)^2.$
AnswerLet P(k) be true for some positive integer k, i.e., $1+2+3+...+\text{k}<\frac{1}{8}(2\text{k}+1)^2\ \ ...(1)$ We shall now prove that P(k + 1) is true whenever P(k) is true. Consider inequality (1)$1+2+3+...+\text{k}<\frac{1}{8}(2\text{k}+1)^2$
Adding (k + 1) on both the sides of the inequality, we have, $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}(2\text{k}+1)^2+(\text{k+1})$ $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}\{(2\text{k}+1)^2+8(\text{k+1})\}$ $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}\{4\text{k}^2+4\text{k}+1+8\text{k}+8\}$ $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}\{4\text{k}^2+12\text{k}+9\}$ $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}(2\text{k+3})^2$ $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}\{2(\text{k+1})+1\}^2$ Hence, $(1+2+...+\text{k})+(\text{k+1})<\frac{1}{8}(2\text{k}+1)^2+(\text{k+1})$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 55 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1+3+3^2+....+3^{\text{n}-1}=\frac{(3^{\text{n}}-1)}{2}.$
AnswerLet the given statement be P(n), i.e., $\text{P(n)}:1+3+3^2+....+3^{\text{n}-1}=\frac{(3^{\text{n}}-1)}{2}$ For n = 1, we have$\text{P(1)} = \frac{(3^1-1)}{2}=\frac{3-1}{2}=\frac{2}{2}=1$
which is true. Let P(k) be true for some positive integer k, i.e., $1+3+3^2+....+3^{\text{k-1}}=\frac{3^{\text{k}}-1}{2}\ \ ....(\text{i})$ We shall now prove that P(k + 1) is true. Consider,$1+3+3^2+… +3^{\text{k}-1}+3(\text{k}+1)-1$
$=(1+3+3^2+…+3^{\text{k}-1})+3^\text{k}$
$=\frac{(3^{\text{k}}-1)}{2}+3^{\text{k}}\ \ [\text{Using (i)}]$
$=\frac{(3^{\text{k}}-1)+2.3^{\text{k}}}{2}$
$=\frac{(1+2)3^{\text{k}}-1}{2}$
$=\frac{3.3^{\text{k}}-1}{2}$
$=\frac{3^{\text{k}+1}-1}{2}$
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 65 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3\text{n}-2)(3\text{n}+1)}=\frac{\text{n}}{(3\text{n}+1)}.$
AnswerLet the given statement be P(n), i.e.,
$\text{P(n)}:\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3\text{n}-2)(3\text{n}+1)}=\frac{\text{n}}{(3\text{n}+1)}$
For n = 1, we have
$\text{P(1)}=\frac{1}{1.4}=\frac{1}{3.1+1}=\frac{1}{4}=\frac{1}{1.4},$
which is true.
Let P(k) be true for some positive integer k, i.e.,
$\text{P(k)}:\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3\text{k}-2)(3\text{k}+1)}=\frac{\text{k}}{(3\text{k}+1)}\ ....(1)$
We shall now prove that P(k + 1) is true.
Consider,
$\Big\{\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3\text{k}-2)(3\text{k}+1)}\Big\}+\frac{1}{\{3(\text{k+1})-2\}\{3(\text{k+1})+1\}}$
$=\frac{\text{k}}{3\text{k}+1}+\frac{1}{(3\text{k+1})(3\text{k}+4)}\ \ [\text{Using (1)}]$
$=\frac{1}{(3\text{k+1})}\Big\{\text{k}+\frac{1}{(3\text{k}+4)}\Big\}$
$=\frac{1}{(3\text{k+1})}\Big\{\frac{\text{k}(3\text{k}+4)+1}{(3\text{k}+4)}\Big\}$
$=\frac{1}{(3\text{k+1})}\Big\{\frac{3\text{k}^2+4\text{k}+1}{(3\text{k}+4)}\Big\}$
$=\frac{1}{(3\text{k+1})}\Big\{\frac{3\text{k}^2+3\text{k}+\text{k}+1}{(3\text{k}+4)}\Big\}$
$=\frac{(3\text{k}+1)(\text{k+1})}{(3\text{k+1})(3\text{k+4})}$
$=\frac{(\text{k+1})}{3(\text{k+1})+1}$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 75 Marks
Prove the following by using the principle of mathematical induction for all $n \in N:$
$\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big)...\Big(1+\frac{(2\text{n+1})}{\text{n}^2}\Big)=(\text{n+1})^2.$
AnswerLet the given statement be P(n), i.e.,
$\text{P(n)}:\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big)...\Big(1+\frac{(2\text{n+1})}{\text{n}^2}\Big)=(\text{n+1})^2$
For n = 1, we have
$\text{P}(1):\Big(1+\frac{3}{1}\Big)=4=(1+1)^2=2^2=4,$
which is true.
Let P(k) be true for some positive integer k, i.e.,
$\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big)...\Big(1+\frac{(2\text{k}+1)}{\text{k}^2}\Big)=(\text{k+1})^2\ ...(\text{i})$
We shall now prove that P(k + 1) is true.
Consider,
$\Big[\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big)...\Big(1+\frac{(2\text{k}+1)}{\text{k}^2}\Big)\Big]\Big\{1+\frac{\{2(\text{k+1})+1\}}{(\text{k}+1)^2}\Big\}$
$=(\text{k+1})^2\Big(1+\frac{2(\text{k+1})+1}{(\text{k+1})^2}\Big)\ \ [\text{Using (i)}]$
$=(\text{k}+1)^2\Big[\frac{(\text{k+1})^2+2(\text{k+1})+1}{(\text{k}+1)^2}\Big]$
$= (k + 1)^2 + 2(k + 1) + 1$
$= {(k + 1) + 1}^2$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 85 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.3+2.3^2+3.3^3+...+\text{n}.3^{\text{n}}=\frac{(2\text{n}-1)3^{\text{n+1}}+3}{4}.$
AnswerLet the given statement be P(n), i.e.,
$\text{P(n)}:1.3+2.3^2+3.3^3+...+\text{n}.3^{\text{n}}=\frac{(2\text{n}-1)3^{\text{n+1}}+3}{4}$
For n = 1, we have
$\text{P(1)}:1.3=3=\frac{(2.1-1)3^{1+1}+3}{4}=\frac{3^2+3}{4}=\frac{12}{4}=3,$
which is true.
Let P(k) be true for some positive integer k, i.e.,
$1.3+2.3^2+3.3^3+...+\text{k}3^{\text{k}}=\frac{(2\text{k}-1)3^{\text{k+1}}+3}{4}\ \ .....(\text{i})$
We shall now prove that P(k + 1) is true.
Consider,
$1.3+2.3^2+3.3^3+...+\text{k}3^{\text{k}}+(\text{k+1})3^{\text{k+1}}$
$=(1.3+2.3^2+3.3^3+...+\text{k}.3^{\text{k}})+(\text{k+1})3^{\text{k+1}}$
$=\frac{(2\text{k}-1)3^{\text{k+1}}+3}{4}+(\text{k+1})3^{\text{k+1}}\ \ [\text{Using (i)}]$
$=\frac{(2\text{k}-1)3^{\text{k+1}}+3+4(\text{k+1})3^{\text{k+1}}}{4}$
$=\frac{3^{\text{k+1}}\{2\text{k}-1+4(\text{k+1})\}+3}{4}$
$=\frac{3^{\text{k+1}}\{6\text{k}+3\}+3}{4}$
$=\frac{3^{\text{k+1}}.3\{2\text{k}+1\}+3}{4}$
$=\frac{3^{(\text{k+1})+1}\{2\text{k}+1\}+3}{4}$
$=\frac{\{2(\text{k+1})-1\}3^{(\text{k+1})+1}+3}{4}$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 95 Marks
Prove the following by using the principle of mathematical induction for all $n \in N:$
$(2n + 7) < (n + 3)^2.$
AnswerLet the given statement be P(n), i.e.,
P(n): $(2n + 7) < (n + 3)^2$
It can be observed that P(n) is true for n = 1.
Since $2.1 + 7 = 9 < (1 + 3)^2 = 16,$
which is true.
Let P(k) be true for some positive integer $k, i.e.,$
$(2k + 7) < (k + 3)^2 ......(1)$
We shall now prove that P(k + 1) is true whenever P(k) Is true.
Consider,
${2(k + 1) + 7} = (2k + 7) + 2$
$\therefore$ ${2(k + 1) + 7} = (2k + 7) + 2 < (k + 3)^2 + 2$ [Using (1)]
$2(k + 1) + 7 < k^2 + 6k + 9 + 2$
$2(k + 1) + 7 < k^2 + 6k + 11$
Now, $k^2 + 6k + 11 < k^2 + 8k + 16$
$\therefore$ $2(k + 1) + 7 < (k + 4)^2$
$2(k + 1) + 7 < {(k + 1) + 3}^2$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 105 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^{\text{n-1}}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}.$
AnswerLet the given statement be P(n), i.e., $\text{P(n)}:\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^{\text{n-1}}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}$ For n = 1, we have $\text{P}(1):\text{a}=\frac{\text{a}(\text{r}^1-1)}{(\text{r}-1)}=\text{a},$ which is true. Let P(k) be true for some positive integer k, i.e., $\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^{\text{k-1}}=\frac{\text{a}(\text{r}^\text{k}-1)}{\text{r}-1}\ \ ....(\text{i})$ We shall now prove that P(k + 1) is true. Consider, $\{\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^{\text{k+1}}\}+\text{ar}^{(\text{k+1})-1}$ $=\frac{\text{a}(\text{r}^{\text{k}}-1)}{\text{r}-1}+\text{ar}^{\text{k}}\ \ [\text{Using (i)}]$$=\frac{\text{a(r}^{\text{k}}-1)+\text{ar}^\text{k}(\text{r-1})}{\text{r}-1}$
$=\frac{\text{a}(\text{r}^{\text{k}}-1)+\text{ar}^{\text{k+1}}-\text{ar}^{\text{k}}}{\text{r}-1}$ $=\frac{\text{ar}^{\text{k}}-\text{a}+\text{ar}^{\text{k+1}}-\text{ar}^{\text{k}}}{\text{r}-1}$ $=\frac{\text{ar}^{\text{k+1}}-\text{a}}{\text{r}-1}$ $=\frac{\text{a(r}^{\text{k+1}}-1)}{\text{r}-1}$Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 115 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:
$1^2+3^2+5^2+...+(2\text{n}-1)^2=\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}.$
Answer$\text{Let P(n)}=1^2+3^2+5^2+...+(2\text{n}-1)^2=\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$ For n = 1 $\text{P}(1)=(2\times1-1)^2=\frac{1(2\times1-1)(2\times1+1)}{3}$ ⇒ 1 = 1 $\therefore$ P(1) is true. Now, let P(n) be true for n = k $\text{P(k)}=1^2+3^2+5^2+...+(2\text{k}-1)^2=\frac{\text{k}(2\text{k}-1)(2\text{k}+1)}{3}\ .....(\text{i})$ For n = k + 1 $\text{R.H.S.}=\frac{(\text{k+1})(2\text{k+1})(2\text{k}+3)}{3}$ $\therefore\ \text{L.H.S.}=\frac{\text{k}(2\text{k}-1)(2\text{k+1})}{3}+(2\text{k}+1)^2\ \ [\text{Using (i)}]$ $=(2\text{k}+1)\Big[\frac{\text{k}(2\text{k}+1)}{3}+(2\text{k+1})\Big]$ $=(2\text{k}+1)\Big[\frac{\text{k}^2-\text{k}+6\text{k}+3}{3}\Big]$ $=\frac{(2\text{k}+1)(2\text{k}^2+5\text{k}+3)}{3}$$=\frac{(2\text{k}+1)(\text{k+1})(2\text{k+3})}{3}$
$=\frac{(\text{k}+1)(2\text{k+1})(2\text{k+3})}{3}$
$\therefore$ P(k + 1) is true.
Therefore, P(k) is true.
⇒ P(k + 1) is true.
Hence by Principle of Mathematical Induction, P(n) is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 125 Marks
Use the Principle of Mathematical Induction in the following Exercis.
Prove that $\frac{1}{\text{n}+1}+\frac{1}{\text{n}+2}+\ .....\ +\frac{1}{2\text{n}}>\frac{13}{24},$ for all natural numbers n > 1.
AnswerLet $\text{P(n): }\frac{1}{\text{n}+1}+\frac{1}{\text{n}+2}+\ .....\ +\frac{1}{2\text{n}}>\frac{13}{24},\forall\text{ n}\in\text{N}$
Step 1: $\text{P}(2){:}\ \frac{1}{2+1}+\frac{1}{2+2}>\frac{13}{24}$
$\Rightarrow\frac{1}{3}+\frac{1}{4}>\frac{13}{24}$
$\Rightarrow\frac{7}{12}>\frac{13}{24}$
$\Rightarrow\frac{14}{24}>\frac{13}{24}$
$\Rightarrow\text{P}(2)$ is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N}$
$\Rightarrow\text{P(k): }\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\ .....\ +\frac{1}{2\text{k}}>\frac{13}{24}\ ...(\text{i})$ is true.
Step 3: Now we have to prove $\text{P(k+ 1):}\ \frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\ .....+\ \frac{1}{2\text{k}}+\frac{1}{2(\text{k}+1)}>\frac{13}{24}$ is true.
Adding $\frac{1}{2(\text{k}+1)}$ on both sides of equation (i)
So, $\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\ .....\ +\frac{1}{2\text{k}}+\frac{1}{2(\text{k}+1)}>\frac{13}{24}+\frac{1}{2(\text{k}+1)}>\frac{13}{24}$
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
View full question & answer→Question 135 Marks
Prove the following statement by principle of mathematical induction:
$2 + 4 + 6 + ..... + 2n = n^2 + n$ for all natural numbers n.
AnswerLet $P(n): 2 + 4 + 6 + .... + n^2 + n$, $\forall\text{n}\in\text{N}$
Step 1: $P(1): 2 = 1^2 + 1 = 2.$ So, P(1) is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N}\Rightarrow\text{P(k)}:2 + 4+6+\ ....+2\text{k}=\text{k}^2+\text{k}\ .....(\text{i})$ is true.
Step 3: Now we have to prove $P(k + 1) : 2 + 4 + 6 + .... + 2k + 2(k + 1) = (k + 1)^2 + (k + 1)$ is true.
Adding 2(k + 1) on both sides of equation (i)
$2 + 4 + 6 + ..... + 2k + 2(k + 1) = k^2 + k + 2(k + 1)$
$= k^2+ 2k + 1 + k + 1$
$= (k + 1)^2 + (k + 1)$
P(k + 1) is true.
So, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction.
View full question & answer→Question 145 Marks
Prove the following statement by principle of mathematical induction:
$2^{3n} - 1$ is divisible by $7,$ for all natural numbers n.
AnswerLet $P(n): 2^{3n} - 1$ is divisible by $7.$
Step 1: $P(1): 2^{3.1} - 1 = 8 - 1 = 7$ which is divisible by 7. So P(1) is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N}\Rightarrow\text{P(k): }2^{3\text{k}}-1=7\lambda, \lambda\ \in\text{ N}$
Step 3: Now we have to prove ${\text{P}(\text{k+1): }}2^{3(\text{k}+1)}-1$ is divisible by 7.
${\text{P}(\text{k+1): }}2^{(\text{k}+1)}-1$
$=2^{3\text{k}+3}-1=2^3\cdot2^{3\text{k}}-8+7$
$=8\cdot2^{3\text{k}}-8+7$ (from step 2)
$=8\cdot7\lambda+7$
$=8(8\lambda+1)$ is divisible by 7
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have p(n) is for all n.
View full question & answer→Question 155 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$
AnswerLet P(n): $\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$
For n = 1
$\frac{1}{3.7}=\frac{3}{(7)}$
$\frac{1}{21}=\frac{1}{21}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4k-1)(4k+3)}}=\frac{\text{k}}{3(\text{4k}+3)} \ ...(1)$
We have to show that
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4k-1)(4k+3)}}+\frac{1}{(4\text{k}+3)(4\text{k}+7)}=\frac{(\text{k}+1)}{3\text{(4k}+7)}$
Now,
$\Big\{\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4k-1)(4k+3)}}\Big\}+\frac{1}{(4\text{k}+3)(4\text{k}+7)}$
$=\frac{\text{k}}{3(4\text{k}+3)}+\frac{1}{(4\text{k}+3)(4\text{k}+7)}$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{\text{k}}{3}+\frac{1}{(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{\text{k}(4\text{k}+7)+3}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{4\text{k}^2+7\text{k}+3}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{4\text{k}^2+4\text{k}+3\text{k}+3)}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{(4\text{k}(\text{k}+1)+3(\text{k}+1)}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{(4\text{k}+3)(\text{k}+1)}{3(4\text{k}+7)}\Big]$
$=\frac{(\text{k}+1)}{3(4\text{k}+7)}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N} $ by PMI
View full question & answer→Question 165 Marks
Use the Principle of Mathematical Induction in the following Exercis.
Show that $\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{7\text{n}}{15}$ is a natural number for all $\text{n}\in\text{N}.$
AnswerLet $\text{P(n): }\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{7\text{n}}{15}$ is a natural number, for all $\text{n}\in\text{N}$
$\text{P}(1){:}\ \frac{1^5}{5}+\frac{1^3}{3}+\frac{7(1)}{15}=\frac{3+5+7}{15}=\frac{15}{15}=1,$ which is a natural number.
Hence, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
$\therefore\ \text{P(k): }\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{7\text{k}}{15}$ is natural number ....(i)
Now, we have to prove that P(k + 1) is true.
$\text{P(k+1):}\ \frac{(\text{k}+1)^5}{5}+\frac{(\text{k}+1)^3}{3}+\frac{7(\text{k}+1)}{15}$
$=\frac{\text{k}^5+5\text{k}^4+10\text{k}^3+10\text{k}^2+5\text{k}+1}{5}+\frac{\text{k}^3+1+3\text{k}^2+3\text{k}}{3}+\frac{7\text{k}+7}{15}$
$=\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{7\text{k}}{15}+\frac{5\text{k}^4+10\text{k}^3+10\text{k}^2+5\text{k}+1}{5}+\frac{3\text{k}^2+3\text{k}+1}{3}+\frac{7}{15}$
$=\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{7\text{k}}{15}+\text{k}^4+2\text{k}^3+2\text{k}^2+\text{k}+\text{k}^2+\text{k}+\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$
$=\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{7\text{k}}{15}+\text{k}^4+2\text{k}^3+3\text{k}^2+2\text{k}+1$
Which is a natural number. [Using (i)]
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.
View full question & answer→Question 175 Marks
Prove the following statement by principle of mathematical induction:
$1 + 2 + 2^2 + ...... + 2n = 2^{n+1} - 1$ for all natural numbers.
AnswerLet $P(n): 1 + 2 + 2^2 + ......... + 2n = 2^{n+1}- 1$, for all natural numbers n.
$P(1): 1 = 2^{0+1} - 1 = 2 - 1 = 1$, which is true.
Hence, P(1) is true.
Let us assume that P(n) is true for some natural number n = k
$P(k): l + 2 + 2^2 + .....+ 2^k= 2^{k+1} - l ......(i)$
Now, we have to prove that P(k + 1) is true.
$P(k + 1): 1 + 2 + 2^2+ ....... + 2^k + 2^{k+1}$
$= 2^{k+1} - 1 + 2^{k+1}$ [Usinig (i)]
$= 2.2^{k+1} - 1 = 1$
$= 2^{(k+1)+1} - 1$
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural n.
View full question & answer→Question 185 Marks
Prove the following by the principle of mathematical induction:
$1 + 3 + 3^2 + ... + 3^{\text{n}-1}=\frac{3^\text{n}-1}{2}$
AnswerLet $P(n)$ be the given statement. Now, $\text{P(n)}=1 + 3 + 3^2 + ... + 3^{\text{n}-1}=\frac{3^\text{n}-1}{2}$
Step 1:
$\text{P(1)}=1 =\frac{3^1-1}{2}=\frac{2}{2}=1$ Hence, $P(1)$ is true.
Step 2:
Let $P(m)$ is true. Then, $1 + 3 + 3^2 + ... + 3^{\text{m}-1}=\frac{3^\text{m}-1}{2}$
We shall prove that $P(m + 1)$ is true.
That is, $1 + 3 + 3^2 + ... + 3^{\text{m}}=\frac{3^\text{m+1}-1}{2}$
Now, we have: $1 + 3 + 3^2 + ... + 3^{\text{m-1}}=\frac{3^\text{m}-1}{2}$
$\Rightarrow1 + 3 + 3^2 + ... + 3^{\text{m-1}}+3\text{m}=\frac{3^\text{m}-1}{2}+3\text{m}$ [Adding $3^m$ to both sides]
$\Rightarrow1 + 3 + 3^2 + ... +3\text{m}=\frac{3^\text{m}-1+2\times3^\text{m}}{2}=\frac{3^\text{m}(1+2)-1}{2}=\frac{3^\text{m+1}-1}{2}$
Hence, P(m + 1) is true.
By the principle of mathematical induction, $P(n)$ is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 195 Marks
Prove the following statement by principle of mathematical induction:
$3^{2n} - 1$ is divisible by $8,$ for all natural numbers $n.$
AnswerLet $P(n): 3^{2n} - 1$
Step 1: $P(1): 3^2 - 1 = 9 - 1 = 8$ which is divisible by $8$
So, $P(1)$ is true.
Step 2: Assume $P(k)$ is true some $\text{k }\in\text{ N}\Rightarrow\text{P(k): }3^{2\text{k}}-1=8\lambda,\lambda\ \in\text{ N}$
Step 3: Now we have to prove $P(k + 1): 3^{2(k+1)} - 1$ is divisible by $8.$
${\text{P}(\text{k+1): }}3^{2(\text{k}+1)}-1$
$=3^{2\text{k}+2}-1=3^2\cdot3^{2\text{k}}-9+8=9(3^{2\text{k}}-1)+8$
$=9\cdot8\lambda+8$ (from step 2)
$=8\big[9\lambda+1\big]$ is divisible by $8$.
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence, $P(k + 1)$ is true whenever $P(k)b$ is true.
Therefore by the principle of mathematical induction we have $P(n)$ is true for all $n.$
View full question & answer→Question 205 Marks
Prove that $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+...+\cos(\alpha+(\text{n}-1)\beta)\\=\frac{\cos\Big\{\alpha+\big(\frac{\text{n}-1}{2}\big)\beta\Big\}\sin\big(\frac{\text{n}\beta}{2}\big)}{\sin\frac{\beta}{2}}$ For all $\text{n}\in\text{N}.$
AnswerLet
P(n): $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+...+\cos(\alpha+(\text{n}-1)\beta]\\=\frac{\cos\Big\{\alpha+\big(\frac{\text{n}-1}{2}\big)\beta\Big\}\sin\big(\frac{\text{n}\beta}{2}\big)}{\sin\frac{\beta}{2}}$ For all $\text{n}\in\text{N}.$
Step 1: For n = 1
$\text{LHS }=\cos[\alpha(1-1)\beta]=\cos\alpha$
$\text{RHS }=\frac{\cos\Big\{\text{a}+\big(\frac{1-1}{2}\big)\beta\Big\}\sin\big(\frac{\beta}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}=\cos\alpha$
As, LHS = RHS
So, it is true for n = 1.
Step II: For n = k
Let p(k): $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ...\ +\cos[\alpha+(\text{k}-1)\beta]=\frac{\cos\big\{\alpha+\big(\frac{\text{k}-1}{2}\big)\beta\big\}\sin\big(\frac{\text{k}\beta}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}$ be true
Step III: For n = k + 1,
$\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+....+\cos(\alpha+(\text{k}-1)\beta]+\cos[\alpha+(\text{k}+1-1\beta)$
$=\frac{\cos\bigg\{\alpha+\Big(\frac{\text{k}-1}{2}\Big)\beta\sin\Big(\frac{\text{k}\beta}{2}\Big)}{\sin\Big(\frac{\beta}{2}\Big)}+\cos(\text{a}+\text{k}\beta)$
$=\frac{\cos\bigg\{\alpha+\Big(\frac{\text{k}-1}{2}\Big)\beta\bigg\}\sin\Big(\frac{\text{k}\beta}{2}\Big)+\sin\big(\frac{\beta}{2}\big)\cos(\text{a}+\text{k}\beta)}{\sin\frac{\beta}{2}}$
$=\frac{\sin\big(\alpha+\text{k}\beta-\frac{\beta}{2}\big)-\sin\big(\alpha-\frac{\beta}{2}\big)+\sin\big(\text{a}+\text{k}\beta+\frac{\beta}{2}\big)-\sin\big(\text{a}+\text{k}\beta-\frac{\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$=\frac{-\sin\big(\alpha-\frac{\beta}{2}\big)\sin\big(\frac{\text{k}\beta+\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$=\frac{2\cos\big(\frac{2\text{a}+\text{k}\beta}{2}\big)\sin\big(\frac{\text{k}\beta+\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$=\frac{2\cos\big(\frac{2\text{a}+\text{k}\beta}{2}\big)\sin\big(\frac{(\text{k}+1)+\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$\text{RHS}=\frac{\cos\Big\{\alpha+\big(\frac{\text{k}+1-1}{2}\big)\beta\Big\}\sin\big(\frac{(\text{k}+1\beta)}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}$
$=\frac{\cos\big(\alpha+\frac{\text{k}\beta}{2}\big)\sin\big(\frac{(\text{k}+1\beta)}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}$
As, LHS = RHS
So, it is also true for n = k + 1.
View full question & answer→Question 215 Marks
Prove the following by the principle of mathematical induction:
$1.3 + 3.5 + 5.7 + ... + (2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$
AnswerLet P(n): $1.3 + 3.5 + 5.7 + ... +(2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$
for n = 1
$1.3=\frac{1(4+6-1)}{3}$
3 = 3
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$1.3 + 3.5 + 5.7 + ... +(2\text{k} - 1)(2\text{k} + 1)= \frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3} \ ...(1)$
We have to show that
$1.3 + 3.5 + 5.7 + ... +(2\text{k} - 1)(2\text{k} + 1)+(2\text{k} + 1)(2\text{k} + 3) $
$=\frac{(\text{k}+1)\big[4(\text{k} + 1)^2+6(\text{k+1})-1\big]}{3}$
Now,
${1.3 + 3.5 + 5.7 + ... + (2\text{k} - 2)(2\text{k} + 1)} + (2\text{k} + 1)(2\text{k} + 3)$
$=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}+(2\text{k}+1)(2\text{k}+3)$ [Using equation (1)]
$=\frac{\text{k}(4\text{k}^2+6\text{k}-1)+3(4\text{k}^2+6\text{k}+2\text{k}+3)}{3}$
$=\frac{4\text{k}^3+6\text{k}-\text{k}+12\text{k}^2+18\text{k}+6\text{k}+9}{3}$
$=\frac{4\text{k}^3+18\text{k}^2+23\text{k}+9}{3}$
$=\frac{4\text{k}^3+4\text{k}^2+14\text{k}^2+14\text{k}+9\text{k}+9}{3}$
$=\frac{(\text{k}+1)(4\text{k}^2+8\text{k}+4+6\text{k}+6-1)}{3}$
$=\frac{(\text{k}+1)\big[(4(\text{k}+1)^2+6(\text{k}+1)-1\big]}{3}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 225 Marks
Prove that $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$ for all $\text{n}\in\text{N}.$
AnswerP(n): $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$
For n = 1
$\frac{2!}{2^2.1}\leq\frac{1}{\sqrt{4}}$
$=\frac{1}{2}\leq\frac{1}{2}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\frac{(2\text{k})!}{2^{2\text{k}}(\text{k}!)^2}\leq\frac{1}{\sqrt{3\text{k}+1}} \ ...(1)$
We have to show that,
$\frac{2(\text{k+1})!}{2^{2(\text{k+1})}\big[(\text{k+1})!\big]^2}\leq\frac{1}{\sqrt{3\text{k}+4}}$
Now,
$\frac{2(\text{k+1})!}{2^{2(\text{k+1})}\big[(\text{k+1})!\big]^2}$
$=\frac{(2\text{k+2})!}{2^{2\text{k}}.2^{2}(\text{k+1})!{(\text{k+1})!}}$
$=\frac{(2\text{k+2})(2\text{k+1})(2\text{k})!}{4.2^{2}(\text{k+1})(\text{k}!){(\text{k+1})(\text{k}!)}}$
$=\frac{2(\text{k+2})(2\text{k+1})(2\text{k})!}{4.(\text{k+1})^2.2^{2\text{k}}.(\text{k}!)}^2$
$\leq\frac{2(2\text{k+1})}{4(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+1}}$
$\leq\frac{(2\text{k+1})}{2(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+1}}$
$\leq\frac{(2\text{k+1})}{2(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+3+1}}$
$\leq\frac{1}{\sqrt{3\text{k}+4}} \ \begin{bmatrix}\text{Since,} \ 2 \text{k}+2<2\text{k}+2\\\ 3\text{k}+1\leq 3\text{k}+4\end{bmatrix}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 235 Marks
Prove the following by the principle of mathematical induction:
1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! - 1 for all $\text{n}\in\text{N}$
AnswerConsider equation
1 × 1! + 2 × 2! + 3 × 3! + ... + n × n!
Lets take (n + 1)! -n! = n!(n + 1 - 1) = n × n!
Now substitue n = 1, 2, 3, 4 ... n in above equation we get
2! - 1! = 1 × 1!
3! - 2! = 2 × 2!
4! - 3! = 3 × 3!
.....
(n + 1)! -n! = n × n!
Adding all the above terms gives
1 × 1! + 2 × 2! + 3 × 3! + ... n × n! = 2! - 1! + 3! - 2! + 4! - 3! ... (n + 1)! - n!
1 × 1! + 2 × 2! + 3 × 3! + ... n × n! = (n + 1)! - 1
View full question & answer→Question 245 Marks
A sequence $a_1, a_2, a_3, ...$ is defined by letting $a_1 = 3$ and $a_k = 7a_k - 1$ for all natural numbers $\text{k}\geq2.$ Show that $a_n = 3.7^{n-1}$ ^for all $\text{n}\in\text{N}.$
AnswerLet $p(n)$ be the statement given by
$P(n): a_n = 3 \times 7^{n-1}$ for all $\text{n}\in\text{N}.$
Step $1:$
$p(2): a_2 = 3 \times 7^{2-1} = 21$
Given that $a_k = 7a_{k-1}$_ for all natural numbers $\text{k}\geq2$
$a_2 = 7a_1 = 7 \times 3 = 21$
$\therefore p(2)$ is true.
Step $2:$
Let $p(m)$ is true. Then,
$a_m = 3 \times _7^{m-1} ... (1)$
We have to prove that $p(m + 1)$ is true.
$a_{m+1} = 7a_m$
$a_{m+1} = 7 \times a_m$
$a_{m+1} = 7^1 \times 3 \times 7^{m-1} ... [$From $(1)]$
$a_{m+1} = 3 \times 7^{m-1+1}$
$a_{m+1} = 3 \times 7^m$
$\Rightarrow p(m + 1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 255 Marks
Prove the following by the principle of mathematical induction:
$5^{2n+2} + 24n - 25$ is divisible by $576$ for all $\text{n}\in\text{N}$
AnswerLet $p(n): 5^{2n+2} + 24n - 25$ is divisible by $576$ for all $576$
For $n = 1$
$5^4 - 24 - 25$
$= 625 - 49$
$= 576$
Which is divisible by $576$
$\Rightarrow p(n)$ is true for $n = 1$
Let $p(n)$ is true for $n = k,$ so
$5^{2k+2} - 24k - 25$ is divisible by $576$
$5^{2\text{k}+2} - 24\text{k} - 25=576\lambda \ ...(1)$
We have to show that,
$5^{(2k+2)+2} - 24(k+1) - 25$ is divisible by $576$
$5^{(2\text{k}+2)+2} - 24(\text{k}+1) - 25=576\mu$
Now,
$5^{(2\text{k}+2)+2} - 24(\text{k}+1) - 25$
$=5^{(2\text{k}+2)}.5^2 - 24\text{k}-24-25$
$(576\lambda+24\text{k}+25)25-24\text{k}-49 [$Using equation $(1)]$
$=25.576\lambda+600\text{k}+625-24\text{k}-49$
$=25.576\lambda+576\text{k}+576$
$=576(25\lambda+\text{k}+1)$
$=576\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 265 Marks
Prove the following by the principle of mathematical induction:
$1 + 3 + 5 + ... + (2n - 1) = n^2$ i.e., the sum of first n odd natural numbers is $n^2.$
AnswerLet $P(n): 1 + 3 + 5 + ... + (2n - 1) = n^2$
For $n = 1$
$P(1): 1 = 1^2$
$1 = 1$
$\Rightarrow P(n)$ is true for $n = 1$
Let $P(n)$ is true for $n = k,$ so
$P(k): 1 + 3 + 5 + ... + (2k - 1) = k^2 ... (1)$
We have to show that
$1 + 3 + 5 + ... + (2k - 1) + 2(k + 1) - 1= (k + 1)^2$
Now,
$\{1 + 3 + 5 + ... + (2k - 1)\} + 2(k + 1)$
$= k^2 + 2(k + 1) [$Using equation $(1)]$
$= k^2 + 2k + 1$
$= (k + 1)^2$
$\Rightarrow P(n)$ is true for $n = k + 1$
$\Rightarrow P(n)$ is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 275 Marks
Prove that $\frac{1}{\text{n}+1}+\frac{1}{\text{n}+2}+...+\frac{1}{2\text{n}}>\frac{13}{24},$ For all natural numbers n > 1.
Answer$\frac{1}{\text{n}+1}+\frac{1}{\text{n}+2}+...+\frac{1}{2\text{n}}>\frac{13}{24},$
Using induction we first show this is true for n = 2.
$\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}>\frac{13}{24}$ (True)
Now lets assume it is true for some n = k,
$\text{S}_\text{k}=\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+...+\frac{1}{2\text{k}}>\frac{13}{24}$
Finally we need to prove that this implies it is also true for n = k + 1:
$\text{S}_\text{k+1}=\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}+2}$
$=\frac{-1}{\text{k}+1}+\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$
$=\frac{-1}{\text{k}+1}+\text{S}_\text{k}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$
$=\text{S}_\text{k}+\frac{1}{2(2\text{k}+1)(\text{k}+1)}$
$>\text{S}_\text{k}$
$\therefore\text{S}_\text{k+1}>\frac{13}{24}$
View full question & answer→Question 285 Marks
Use the Principle of Mathematical Induction in the following Exercis.
We have a sequence $d_1, d_2, d_3 .......$ defined by letting $d_1 = 2$ and $\text{d}_\text{k}=\frac{\text{d}_{\text{k}-1}}{\text{k}},$ for all natural numbers $\text{k}\geq2.$ show that $\text{d}_\text{n}=\frac{2}{\text{n}!},$ for all $\text{n}\in\text{N}$
AnswerWe have a sequence $d_1, d_2, d_3 .......$ defined by letting $d_1 = 2$ and $\text{d}_\text{k}=\frac{\text{d}_{\text{k}-1}}{\text{k}},$
Let $\text{P(n)}:\text{d}_\text{n}=\frac{2}{\text{n}!}\forall\text{n}\in\text{N}$
$\text{P}(2):\text{d}_2=\frac{2}{2!}=\frac{2}{2\times1}=1$
Also, $d_1 = 2$ and $\text{d}_\text{k}=\frac{\text{d}_{\text{k}-1}}{\text{k}}$
$\text{d}_2=\frac{\text{d}_1}{2}=\frac{2}{2}=1$
Hence, $P(2)$ is true.
Now, let us assume that $P(n)$ in truw for some natural number $n = m.$
$\therefore\ \text{P(m)}:\text{d}_\text{m}=\frac{2}{\text{m}!}\ ...(\text{i})$
Now, to prove that $P(m + 1)$ is true, we have to show that
$\text{P}(\text{m}+1):\text{d}_{\text{m}+1}=\frac{2}{(\text{m}+1)!}$
$\text{d}_{\text{m}+1}=\frac{\text{d}_{\text{m}+1-1}}{\text{m}+1}=\frac{\text{d}_\text{m}}{\text{m}+1}=\frac{2}{\text{m}!(\text{m}+1)}=\frac{2}{(\text{m}+1)!}$
Hence, $P(m + 1)$ is true whenever $P(m)$ is true.
So, by the principle of mathematical induction $P(n)$ is true for any natural number $n.$
View full question & answer→Question 295 Marks
Prove the following by the principle of mathematical induction:
$3^{2n} + 7$ is divisible by $8$ for all $\text{n}\in\text{N}$
AnswerLet $p(n): 3^{2n} + 7$ is divisible by 8 for all $\text{n}\in\text{N}$
For $n = 1$
$3^2 + 7 = 16$
Which is divisible by $8$
$\Rightarrow p(n)$ is true for $n = 1$
Let $p(n)$ is true for $n = k,$ so
$3^{2k} + 7$ is divisible by $8$
$\Rightarrow3^{2\text{k}} + 7 =8\lambda \ ...(1)$
We have to show that,
$3^{2(k+1)} + 7$ is divisible by $8$
$3^{(2\text{k+1})} + 7 =8\mu$
Now,
$3^{2(k+1)} + 7$
$= 3^{2k}.3^2 + 7$
$= 9.3^{2k} + 7$
$=9.(8\lambda-7)+7$
$=72\lambda-56$
$=8(9\lambda-7)$
$=8\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 305 Marks
Prove the following by the principle of mathematical induction:
$1.3 + 2.4 + 3.5 + ... + \text{n}(\text{n} + 2) $
$=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+7)$
AnswerLet $1.3 + 2.4 + 3.5 + ... + \text{n}(\text{n} + 2) $
$=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+7)$
for n = 1
$1.3=\frac{1}{6}.1.(2)(9)$
3 = 3
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$1.3 + 2.4 + 3.5 + ... + \text{k}(\text{k} + 2) =\frac{1}{6}\text{k}(\text{k}+1)(2\text{k}+7) \ ...(1)$
We have to show that
$1.3 + 2.4 + 3.5 + ... + \text{k}(\text{k} + 2)+(\text{k} + 1)(\text{k} + 3)$
$ =\frac{\text{k}+1}{6}(\text{k}+2)(2\text{k}+9)$
Now,
{1.3 + 2.4 + 3.5 + ... + k(k + 2)} + (k + 1)(k + 3)
$ =\frac{1}{6}\text{k}(\text{k}+1)(2\text{k}+7)+(\text{k}+1)(\text{k}+3)$ [Using equation (1)]
$(\text{k} + 1)\Big[\frac{\text{k}(2\text{k}+7)}{6}+\frac{\text{k}+3}{1}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}^2+7\text{k}+6\text{k}+18}{6}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}^2+13\text{k}+18}{6}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}^2+4\text{k}+9\text{k}+18}{6}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}+(\text{k}+2)+9(\text{k}+2)}{6}\Big]$
$(\text{k} + 1)\Big[\frac{(2\text{k}+9)(\text{k}+2)}{6}\Big]$
$\frac{1}{6}(\text{k}+1)(\text{k}+2)(2\text{k}+9)$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 315 Marks
Use the Principle of Mathematical Induction in the following Exercis.
A sequence $a_1, a_{2,} a_3 ......$ is defined by letting $a_1 = 3$ and $a_k = 7a_{k-1}$ for all natural numbers $\text{k}\geq2.$ Show that an $= 3.7^{n-1}$ for all natural numbers.
AnswerWe have a sequence $a_x, a_2, a_3 .......$ defined by letting $a = 3$ and $a_k = 7a_{k-1}$ for all natural numbers
$\text{k}\geq2.$
Let $P(n): a_n = 3 \times 7^{n-1}$ for all natural numbers.
For $n = 2, a_2 = 3 \times 7^{2-1} = 3 \times 7^1 = 21$
Also, $a_1 = 3, ak = 7a_{k-1}$
$\Rightarrow a_2 = 7 \times a_1 = 7 \times 3 = 21$
Thus, P(2) ius true.
Now, let us assume that P(n) is true for some natural number n = m.
$\therefore$ P(m): am $= 3 \times 7^{m-1}$
Now, to prove that P(m + 1) is true, we have to show that,
$P(m + 1): a_{m+1} = 3 \times 7^{m+1-1}$
$\Rightarrow a_{m+1} = 7 \times a_{m+1-1} (as a_k = 7a_{k-1})$
$\Rightarrow a_{m+1} = 7 \times a_m$
$\Rightarrow a_{m+1} = 7 \times 3 \times 7^{m-1}$
$\Rightarrow a_{m+1} = 3 \times 7^{m-1+1}$
Hence, P(m + 1) is true whenever P(m) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.
View full question & answer→Question 325 Marks
Prove that $\Big(1-\frac{1}{2^2}\Big)\Big(1-\frac{1}{3^2}\Big)\Big(1-\frac{1}{4^2}\Big)...\Big(1-\frac{1}{\text{n}^2}\Big)=\frac{\text{n}+1}{2\text{n}}$ for all natural numbers, $\text{n}\geq2.$
Answer$\Big(1-\frac{1}{2^2}\Big)\Big(1-\frac{1}{3^2}\Big)\Big(1-\frac{1}{4^2}\Big)...\Big(1-\frac{1}{\text{n}^2}\Big)$
Above can be written as
$=\Big(\frac{2^2-1}{2^2}\Big)\Big(\frac{3^2-1}{3^2}\Big)\Big(\frac{4^2-1}{4^2}\Big)...\Big(\frac{\text{n}^2-1}{\text{n}^2}\Big)$
$=\Big(\frac{(2+1)(2-1)}{2^2}\Big)\Big(\frac{(3+1)(3-1)}{3^2}\Big)\Big(\frac{(4+ 1)(4-1)}{4^2}\Big)...\Big(\frac{(\text{n}+1)(\text{n}-1)}{\text{n}^2}\Big)$
$=\Big(\frac{3.1}{2^2}\Big)\Big(\frac{4.2}{3^2}\Big)\Big(\frac{5.3}{4^2}\Big)...\Big(\frac{(\text{n}+1)(\text{n}-1)}{\text{n}^2}\Big)$
In the above product, there are two series in numerator 3.4.5...(n + 1) and 1.2.3...(n - 1). All numbers from 3 to (n-1) are repeated twice and 1,2,n are appeared once in numerator. So after cancelling like terms we get
$=\frac{\text{n}+1}{2\text{n}}$
View full question & answer→Question 335 Marks
Prove the following by the principle of mathematical induction:
$1 + 2 + 3 + ... + \text{n}=\frac{\text{n}(\text{n}+1)}{2}$ i.e, the sum of the first n natural numbers is $\frac{\text{n}(\text{n}+1)}{2}.$
AnswerLet P(n): $1 + 2 + 3 + ... + \text{n}=\frac{\text{n}(\text{n}+1)}{2}$
For n = 1,
LHS of P(n) = 1
RHS of P(n) $=\frac{\text{1}(\text{1}+1)}{2}1=1$
Since, LHS = RHS
⇒ P(n) is true for n = 1
Let P(n) be true for n = k, so
$1 + 2 + 3 + ... + \text{k}=\frac{\text{k}(\text{k}+1)}{2}...(1)$
Now,
$ (1 + 2 + 3 + ... + \text{k}) + (\text{k} + 1)$
$=\frac{\text{k}(\text{k}+1)}{2}+(\text{k}+1)$
$=(\text{k}+1)\Big(\frac{\text{k}}{2}+1\Big)$
$=\frac{(\text{k}+1)(\text{k}+2)}{2}$
$=\frac{(\text{k}+1)[(\text{k}+1)+1]}{2}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$
So, by the principle of mathematical induction
P(n): $ 1 + 2 + 3 + ... + \text{n}=\frac{\text{n}(\text{n}+1)}{2}$ is ture for all $\text{n}\in\text{N}$
View full question & answer→Question 345 Marks
Prove the following by the principle of mathematical induction:
$11^{n+2} + 122^{n+1} $is divisible of $133$ for all $\text{n}\in\text{N}$
AnswerLet $p(n): 11^{n+2} + 122^{n+1} $is divisible of 133
For $n = 1$
$11^3 + 12^3$
$= 1331 + 1728$
$= 3059$
It is divisible of 133
$\Rightarrow p ( n )$ is true for $n =1$
Let $p(n)$ is true for $n=k$, so
$11^{k+2}+12^{2 k+1}$ is divisible of 133
$11^{k+2}+12^{2 k+1}=133 \lambda \ldots(1)$
We have to show that, $1^{k+3}+12^{2 k+3}$ is divisible of 133
Now,
$11^{\text{k}+2}.11+12^{2\text{k}+1}.12^2$
$= (133\lambda-12^{2\text{k}+1})11+12^{2\text{k}+1}.144$
$=11.133\lambda+11.12^{2\text{k}+1}+144.12^{2\text{k}+1}$
$=11.133\lambda+133.12^{2\text{k}+1}$
$=133(11\lambda+12^{2\text{k}+1})$
$=133\mu$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 355 Marks
Prove the following by the principle of mathematical induction:
$1^2 + 2^2 + 3^2 +...+ \text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
AnswerLet P(n): $1^2 + 2^2 + 3^2 + ... +\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
For n = 1
$\text{P}(1):1=\frac{1(1+1)(2+1)}{6}$
$1=1$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
P(k): $1^2 + 2^2 + 3^2 + .... + \text{k}^2=\frac{\text{k}(\text{k}+1)(2\text{k}+1)}{6}\ \cdots(1)$
We have to show that P(n) is true for $n = k + 1$
$\Rightarrow 1^2 + 2^2 + 3^2 + .... + k^2 + (k + 1)^2$
$=\frac{(\text{k}+1)(\text{k}+2)(2\text{k}+3)}{6}$
So, $1^2 + 2^2 + 3^2 + .... + k^2 + (k + 1)^2$
$=\frac{\text{k}(\text{k}+1)(2\text{k}+1)}{6}+(\text{k}+1)^2$ [Using equation (1)]
$=(\text{k}+1)\Big[\frac{2\text{k}^2+\text{k}}{6}+\frac{(\text{k}+1)}{1}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k}^2+\text{k}+6\text{k}+6}{6}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k}^2+7\text{k}+6}{6}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k}^2+4\text{k}+3\text{k}+6}{6}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k(k+2)+3(k+2)}}{6}\Big]$
$=\frac{(\text{k}+1)(2\text{k+3})(\text{k}+2)}{6}$
$\Rightarrow P(n)$ is true for $n = k + 1$
$\Rightarrow P(n)$ is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 365 Marks
Prove that number of subsets of a set containing n distinct elements is $2^{ n }$, for all $n \in N$.
AnswerLet $P(n)$ : Number of subset of a set containing $n$ distinct elements is 2 , for all $n N$.
For $n =1$, consider set $A =\{1\}$. So, set of subsets is $\{\{1\}, \varnothing\}$, which contains $2^1$ elements.
So, $P(1)$ is true.
Let us assume that $P(n)$ is true, for some natural number $n=k$.
$P(k)$ : Number of subsets of a set containing $k$ distinct elements is $2^k$ To prove that $P(k+1)$ is true, we have to show that,
$P(k+1)$ : Number of subsets of a set containing $(k+1)$ distinct elements is $2^{k+1}$
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing $(k+1)$ distinct elements $=2 \times 2^k=2^{k+1}$
So, $P(k+1)$ is true. Hence, $P(n)$ is true.
View full question & answer→Question 375 Marks
Prove that the number of subsets of a set containing n distinct elements is $2^n $for all $\text{n}\in\text{N}.$
AnswerLet $p(n)$ be the statement given by
$P(n)$ : The number of subsets of a set containing $n$ distinct element is is $2^n$ for all $n \in N$.
Step 1:
$P(1): 2^1=2$
For any set A containing 1 element, empty set and set A are two sets always subsets of A .
$\therefore P (1)$ is true.
Step 2:
Let $p(m)$ is true. Then,
A set containing $m$ distinct elements has $2^m$ subsets ...
We have to prove that $p(m+1)$ is true.
Let the set A has $(m+1)$ elements.
$A=\{1,2 \ldots, m, m+1\}$
$A=\{1,2 \ldots, m\} \cup\{m+1\}$
Now using (1) we can say that $\{1,2, \ldots, m\}$ being $m$ elements has $2^m$ subsets
For $\{m+1\}$, empty set and itself $\{m+1\}$ are subsets
So, $\{m+1\}$ has 2 subsets
$\Rightarrow$ Set $A$ has $2^m+2$ subsets
$\Rightarrow$ Set $A$ has $2^{m+1}$ subsets
$\Rightarrow p(m+1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $n \in N$.
View full question & answer→Question 385 Marks
Use the Principle of Mathematical Induction in the following Exercis.
A sequence $b_0, b_1, b_2, \ldots .$. is defined by letting $b_0=5$ and $b_k=4+b_{k-1}$ for all natural numbers $k$. Show that $b_n=5+4 n$ for all natural number n using mathematical induction.
AnswerWe have $b _0=5$ and $b _{ k }=4+ b _{ k -1}$
$\Rightarrow b_0=5, b_1=4+b_0=4+5=9$ and $b_2=4+b_1=4+9=13$
Let $P(n): b_n=5+4 n$
Step 1: $P(1): b_1=5+4=9 \Rightarrow 9=9$. So $P(1)$ is true.
Step 2: Assume $P(k)$ is true for some $k \in N \Rightarrow P(k)$ : $b_k=5+4 k$ is true.
Step 3: Now we have to prove $P(k+1)$ : $b_{k+1}=5+4(k+1)$ is true.
As $b_k=4+b_{k-1}$ we have
$b_{k+1}=4+b_{k+1-1}=4+b_k=4+5+4 k=5+4(k+1)$
$\Rightarrow P(k+1)$ is true.
Hence, $P(k+1)$ is true whenever $P(k)$ is true.
Therefore by the principle of mathematical induction we have $P(n)$ is true for all $n$.
View full question & answer→Question 395 Marks
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.
AnswerLet P(n) be the statement 3n < n!
For n = 1,
3n = 3 × 1 = 3
n! = 1! = 1
Now, 3 > 1
So, P(1) is not true.
For n = 2,
3n = 3 × 2 = 6
n! = 2! = 2
Now, 6 > 2
So, P(2) is not true.
For n = 3,
3n = 3 × 3 = 9
n! = 3! = 6
Now, 9 > 6
So, P(3) is not true.
For n = 4,
3n = 3 × 4 = 12
n! = 4! = 24
Now, 12 < 24
So, P(4) is true.
For n = 5,
3n = 3 × 5 = 15
n! = 5! = 120
Now, 15 < 120
So, P(5) is true.
Similarly, it can be verified that 3n < n! for n = 6, 7, 8, ...
Thus, the statement P(n): 3n < n! is true for all n ≥ 4 but P(1), P(2) and P(3) are not true.
View full question & answer→Question 405 Marks
Prove the following by the principle of mathematical induction:
$2.7^n + 3.5^{n-3}.3^{n-1}$ is divisible of 24 for all $\text{n}\in\text{N}$
AnswerLet P(n) be the given statement.
Now,
$P(n): 2.7^n+ 3.5^n - 5$ is divisible by 24.
Step 1:
$P(1): 2.7^1 + 3.5^1 - 5 = 24$
It is divisible by 24.
Thus, P(1) is true.
Step 2:
Let P(m) be true.
Then,
$2.7^m + 3.5^m - 5$ is divisible by 24.
Suppose:
$2.7^\text{m}+3.5\text{m}-5=24\lambda \ ...(1)$
We need to show that P(m + 1) is true whenever P(m) is true.
Now,
$\text{P}(\text{m}+1)=2.7^{\text{m}+1}+3.5^{\text{m}+1}-5$
$=2.7^{\text{m}+1}+(24\lambda+5-2.7^\text{m})5-5$
$=(2.7^{\text{m}+1}+120\lambda+25-10.7^\text{m}5-5)$
$=2.7^{\text{m}}.7-10.7^\text{m}+120\lambda+24-4$
$=7^\text{m}(14-10)+120\lambda+24-4$
$=7^\text{m}.4+120\lambda+24-4$
$=4(7^\text{m}-1)+24(5\lambda+1)$ $\big[$Since $7^m - 1$ is a multiple of 6 for all $\text{n}\in\text{N},7^\text{m} - 1 = \mu\big]$
$=4\times6\mu+24(5\lambda+1)$
$=24(\mu+5\lambda+1)$
It is a multiple of 24.
Thus, P(m + 1) is true.
By the principle of mathematical induction, P(n) is true for $\text{n}\in\text{N}.$
View full question & answer→Question 415 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{n(n+1)}}=\frac{\text{n}}{\text{n}+1}$
AnswerLet P(n): $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{n(n+1)}}=\frac{\text{n}}{\text{n}+1}$
For n = 1
$\text{P}(1):\frac{1}{1.2}=\frac{1}{1+1}$
$\frac{1}{2}=\frac{1}{2}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{k(k+1)}}=\frac{\text{k}}{\text{k}+1}...(1)$
We have to show that
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{k(k+1)}}+\frac{\text{k}}{(\text{k}+1)(\text{k}+2)}=\frac{\text{k}+1}{(\text{k}+2)}$
Now,
$\Big\{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{k(k+1)}}\Big\}+\frac{\text{1}}{(\text{k}+1)(\text{k}+2)}$
$=\frac{\text{k}}{\text{k+1}}+\frac{1}{(\text{k+1)}(\text{k+2)}}$ [Using equation (1)]
$=\frac{1}{\text{k+1}}\Big[\frac{\text{k}(\text{k+2})+1}{(\text{k+2)}}\Big]$
$=\frac{1}{\text{k+1}}\Big[\frac{\text{k}^2+2\text{k}+1}{(\text{k+2)}}\Big]$
$=\frac{1}{\text{k+1}}\Big[\frac{(\text{k}+1)(\text{k}+2)}{(\text{k+2)}}\Big]$
$=\frac{(\text{k}+1)}{(\text{k+2)}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 425 Marks
Give an example of a statement P(n) which is true for all n. Justify your answer.
AnswerThe required statement is,
$\text{p(n): }1+2+3+\ ....... \ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
Justification,
$\text{n}=1,\text{ P(1): }1=\frac{(1+1)}{2}$
Therefore P(1) is true.
Assume $\text{P(k): } 1 + 2 + 3 +\ .....\ +\text{k}=\frac{\text{k}(\text{k}+1)}{2}\ .....(\text{i})$ is true.
Now we have to prove ${\text{P}(\text{k+1): }}1+2+3+\ ....\ +\text{k}+(\text{k}+1)=\frac{(\text{k}+1)(\text{k}+2)}{2}$ is true.
Adding k + 1 on both sides of equation (i) we get,
$1+2+3+\ ....\ +\text{k}+(\text{k}+1)$
$=\frac{\text{k}(\text{k}+1)}{2}+(\text{k}+1)=(\text{k}+1)\Big(\frac{\text{k}}{2}+1\Big)$
$=\frac{(\text{k}+1)(\text{k}+2)}{2}$
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
View full question & answer→Question 435 Marks
Prove the following statement by principle of mathematical induction:
$\sqrt{\text{n}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ ....\ +\frac{1}{\sqrt{\text{n}}},$ for all natural numbers $\text{n}\geq2$
AnswerLet $\text{P(n)}:\sqrt{\text{n}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ ....\ +\frac{1}{\sqrt{\text{n}}},$ for all natural numbers $\text{n}\geq2$
$\text{P(2)}:\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}},$ which is true.
Hence, P(2) is true.
Let us assume that P(n) is true for some natural number n = k
$\therefore\ \text{P(k)}:\sqrt{\text{k}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ .....\ +\frac{1}{\sqrt{\text{k}}}\ ....(\text{i})$
To prove that P(k + 1) is true, we have to show that
$\text{P}{(\text{k}+1)}:\sqrt{\text{k}+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ .....\ +\frac{1}{\sqrt{\text{k}}}+\frac{1}{\sqrt{\text{k}+1}}$
Now, $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ .....\ +\frac{1}{\sqrt{\text{k}}}+\frac{1}{\sqrt{\text{k}+1}}$
$>\sqrt{\text{k}}+\frac{1}{\sqrt{\text{k}+1}}$ [Using(i)]
$>\sqrt{\text{k}+1}$ $\Big(\because\frac{1}{\sqrt{\text{k}+1}}>0\Big)$
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n, $\text{n}\geq2$
View full question & answer→Question 445 Marks
Prove the following statement by principle of mathematical induction:
$4^n- 1$ is divisible by $3,$ for each natural number.
AnswerLet $P(n): 4^n-1$ is divisible by 3 for each natural number $n$.
Now, $P(l): 4^1-1=3$, which is divisible by 3
Hence, $P(l)$ is true.
Let us assume that $P(n)$ is true for some natural number $n=k$.
$P(k): 4^k-1$ is divisible by 3 or $4^k-1=3 m, m \in N \ldots$...(i)
Now, we have to prove that $P(k+1)$ is true.
$P(k+ 1): 4^{k+1} - 1$
$= 4^k- 4 - l$
$= 4(3m + 1) - 1 [Using (i)]$
$= 12 m + 3$
$= 3(4m + 1),$ which is divisible by 3 Thus, $P(k + 1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.
View full question & answer→Question 455 Marks
Prove the following statement by principle of mathematical induction:
$n^3 - 7n + 3$ is divisible by $3$, for all natural numbers n.
AnswerLet $P(n): n^3-7 n+3$ is divisible by 3 , for all natural numbers $n$.
Now $P(1):(1)^3-7(1)+3=-3$, which is divisible by 3 .
Hence, $P (1)$ is true.
Let us assume that $P(n)$ is true for some natural number $n=k$.
$P(k)=K^3-7 k+3$ is divisible by 3 or $K^3-7 k+3=3 m, m \in N$
Now, we have to prove that $P ( k +1)$ is true.
$P(k+1):(k+1)^3-7(k+1)+3$
$=k^3+1+3 k(k+1)-7 k-7+3=k^3-7 k+3+3 k(k+1)-6$
$=3 m+3[k(k+1)-2][\text { sing (i)] $
$=3[m+(k(k+1)-2)]$, which is divisible by 3 Thus, $P(k+1)$ is true whenever $P(k)$ is true.
So, by the principle of mathematical induction $P(n)$ is true for all natural numbers $n$.
View full question & answer→Question 465 Marks
Prove that $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$ for all $\text{n}\in\text{N}.$
AnswerLet P(n): $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$
For n = 1
$\sin \text{x}=\frac{\sin^2\text{x}}{\sin\text{x}}$
$\sin\text{x}=\sin\text{x}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k-1})\text{x}=\frac{\sin ^2\text{kx}}{\sin\text{x}}$
We have to show that,
$\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k-1})\text{x}+\sin(\text{2k+1})\text{x}=\frac{\sin ^2(\text{k+1})\text{x}}{\sin\text{x}}$
Now,
$\Big\{\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k}-1)\text{x}\Big\}+\sin(2\text{k}+1)\text{x}$
$=\frac{\sin^2\text{kx}}{\sin\text{x}}+\frac{\sin(2\text{k}+1)\text{x}}{1}$
$=\frac{\sin^2\text{kx}+{\sin(2\text{k}+1)\text{x}}\sin\text{x}}{\sin\text{x}}$ [Using equation (1)]
$=\frac{2\sin^2\text{kx}+\cos\big[(2\text{k}+1)\text{x}-\text{x}-\big]\big[(2\text{k}+1)\text{x}+\text{x}\big]}{2\sin\text{x}}$
$=\frac{2\sin^2\text{kx}+\cos2\text{kx}-\cos(2\text{kx}+2\text{x})}{2\sin\text{x}}$
$=\frac{1-\cos2\text{kx}+\cos2\text{kx}-\cos2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{1-\cos2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{2\sin^2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{\sin^2\text{x}(\text{k}+1)}{\sin\text{x}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 475 Marks
Prove the following by the principle of mathematical induction:
$2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$ $$
AnswerLet P(n): $2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$
for n = 1
$\text{P}(1)2=\frac{1}{2}.1.(4)$
2 = 2
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$2 + 5 + 8 + 11 + ... +(3\text{k} - 1) =\frac{1}{2}\text{k}(3\text{k}+1) \ ...(1)$
We have to show that
$2 + 5 + 8 + 11 + ... +(3\text{k} - 1)+(3\text{k} + 2) $
$=\frac{1}{2}(\text{k}+1)(3\text{k}+4) $
Now,
${2 + 5 + 8 + 11 + ... + (3\text{k} - 1)} + (3\text{k} + 2)$
$ =\frac{1}{2}\text{k}(3\text{k}+1)(3\text{k}+2)$
$=\frac{3\text{k}^2+\text{k}+2{(3\text{k}+2)}}{2}$
$=\frac{3\text{k}^2+\text{k}+6\text{k}+4}{2}$
$=\frac{3\text{k}^2+7\text{k}+4}{2}$
$=\frac{3\text{k}^2+3\text{k}+4\text{k}+4}{2}$
$\frac{(\text{k}+1)(3\text{k}+4)}{2}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI.
View full question & answer→Question 485 Marks
Prove the following by the principle of mathematical induction:$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
AnswerLet p(n): $\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For n = 1
$\text{a}=\frac{1}{2}[2\text{a}+(1-1)\text{d}]$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, so
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})=\frac{\text{k}}{2}[2\text{a}+(\text{k}-1)\text{d}] \ ...(1)$
We have to show that
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})+(\text{a}+(\text{k})\text{d})=\frac{(\text{k+1})}{2}[2\text{a}+\text{k}\text{d}] $
Now,
$\big\{\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})\big\}+(\text{a}+\text{k}\text{d})$
$\frac{\text{k}}{2}[2\text{a}+(\text{k}-1)\text{d}]+(\text{a}+\text{k}\text{d})$ [Using equation (1)]
$=\frac{2\text{k}\text{a}+\text{k}(\text{k}-1)\text{d}+2(\text{a}+\text{k}\text{d})}{2}$
$=\frac{2\text{k}\text{a}+\text{k}^2\text{d}-\text{k}\text{d}+2\text{a}+2\text{k}\text{d})}{2}$
$=\frac{2\text{k}\text{a}+2\text{a}+\text{k}^2\text{d}+\text{k}\text{d}}{2}$
$=\frac{2\text{a}(\text{k}+1)+\text{d}(\text{k}^2 +\text{k})}{2}$
$=\frac{(\text{k}+1)}{2}[2\text{a}+\text{k}\text{d}]$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 495 Marks
Prove the following by the principle of mathematical induction:
$1 + 2 + 2^2 + ... + 2^n = 2^{n+1} - 1$ for all $\text{n}\in\text{N}.$
AnswerLet p(n) be the statement given by
$p(n): 1 + 2 + 2^2 + ... + 2^n = 2^{n+1} - 1$
Step 1:
$p(1): 1 + 2^1 = 2^{1+1} - 1$
$\Rightarrow 1 + 2 = 4 - 1$
$\Rightarrow 3 = 3$
$\therefore$ p(1) is true.
Step 2:
Let p(m) is true. Then,
$1 + 2 + 2^2 + ... + 2^m = 2^{m+1} - 1 ... (1)$
We have to prove that P(m + 1) is true.
$1 + 2 + 2^2 + ... + 2^{m+1} = 1 + 2 + 2^2 + ... + 2^m + 2^{m+1}$
$= (2^{m+1} - 1) + 2^{m+1} ... [Using (1)]$
$= (2^{m+1} + 2^{m+1}) - 1$
$= 2 \times 2^{m+1} - 1$
$= 2^{m+2} - 1$
$\Rightarrow P(m + 1)$ is true.
Hence by the PMI,l the given result is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 505 Marks
Prove that $\frac{\text{n}^2}{7}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{\text{n}^2}{2}-\frac{37}{210}$ n is a positive integer for all $\text{n}\in\text{N}.$
AnswerLet p(n): $\frac{\text{n}^2}{7}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{\text{n}^2}{2}-\frac{37}{210}$ n is a positive integer
For n = 1
$\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$
$=\frac{30+42+70+105-37}{210}$
$=\frac{247-37}{210}$
It is a positive integer
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}$ k is positive integer
$\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}\text{k}=\lambda$
For n = k + 1,
$\frac{(\text{k+1})^7}{7}+\frac{(\text{k+1})^5}{5}+\frac{(\text{k+1})^3}{3}+\frac{(\text{k+1})^2}{2}-\frac{37}{210}(\text{k+1})$
$=\frac{1}{7}\big[\text{k}^7+7\text{k}^6+21\text{k}^5+35\text{k}^3+21\text{k}^2+7\text{k}+1\big]\\+\frac{1}{5}\big[\text{k}^5+5\text{k}^4+10\text{k}^3+10\text{k}^2+21\text{k}^2+5\text{k}+1\big]\\+\frac{1}{3}\big[\text{k}^3+3\text{k}^2+3\text{k}+1\big]+\frac{1}{2}\big[\text{k}^2+2\text{k}+1\big]-\frac{37\text{k}}{210}-\frac{37}{210}$
$=\Big[\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}\Big]+\Big[\text{k}^6+3\text{k}^5+5\text{k}^4+3\text{k}^2+\text{k}\\\ \ \ \ \ \ +\frac{1}{7}+\text{k}^4+2\text{k}^3+2\text{k}^2+\frac{1}{5}+\text{k}^2+\text{k}+\frac{1}{3}+\text{k}+\frac{1}{2}-\frac{37}{210}\Big]$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+1$
= Positive integer
p(n) is true for n = k + 1
p(n) is true for all by $\text{n}\in\text{N}$ PMI
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