Question 14 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 24 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 34 Marks
Prove the following by using the principle of mathematical induction for all $n ∈ N: n(n + 1)(n + 5)$ 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. ⇒ 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 44 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. $\Rightarrow P(k + 1)$ is true.Hence by Principle of Mathematical Induction, $P(n)$ is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 54 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 64 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 74 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 84 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 94 Marks
Prove the following by using the principle of mathematical induction for all n ∈ 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 104 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 114 Marks
Prove the following by using the principle of mathematical induction for all $n ∈ N: (2n + 7) < (n + 3)^2$.
AnswerLet the given statement be $P(n)$, i.e., $P(n)$ : $(2 n+7)<(n+3)^2$ 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., $(2 k+7)<(k+3)^2 \ldots \ldots . .(1)$ We shall now prove that $P(k+1)$ is true whenever $P(k)$ Is true. Consider, $\{2(k+1)+7\}=(2 k+7)+2$
$\therefore: 2(k+1)+7\}=(2 k+$ 7) $+2<(k+3)^2+2[$ Using (1)]
$2(k+1)+7<k^2+6 k+9+22(k+1)+7<k^2+6 k+11$ Now, $k^2+6 k+11<k^2+8 k+16: 2(k+1)+7<(k+4)^2$
$2(k+1)+7<\{(k+1)+3\}^2$ 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 124 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{n}-1)(3\text{n}+2)}=\frac{\text{n}}{(6\text{n}+4)}.$
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{n}-1)(3\text{n}+2)}=\frac{\text{n}}{(6\text{n}+4)}$
For n = 1, we have$\text{P(1)}=\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10},$
which is true. Let P(k) be true for some positive integer k, i.e.,$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{k}-1)(3\text{k}+2)}=\frac{\text{k}}{(6\text{k}+4)}\ \ ...(\text{i})$
We shall now prove that P(k + 1) is true. Consider,$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{k}-1)(3\text{k}+2)}+\frac{1}{\{3(\text{k}+1)-1\}\{3(\text{k+1})+2\}}$
$=\frac{\text{k}}{(6\text{k}+4)}+\frac{1}{(3\text{k}+3-1)(3\text{k+3+2})}\ \ [\text{Using (i)}]$
$=\frac{\text{k}}{2(3\text{k}+2)}+\frac{1}{(3\text{k}+2)(3\text{k+5})}$
$=\frac{1}{(3\text{k}+2)}\Big(\frac{\text{k}}{2}+\frac{1}{3\text{k}+5}\Big)$
$=\frac{1}{(3\text{k}+2)}\Big(\frac{3\text{k}^2+5\text{k}+2}{2(3\text{k}+5)}\Big)$
$=\frac{1}{(3\text{k}+2)}\Big(\frac{(3\text{k}+2)(\text{k}+1)}{2(3\text{k}+5)}\Big)$
$=\frac{(\text{k+1})}{6\text{k}+10}$
$=\frac{(\text{k+1})}{6(\text{k+1})+4}$
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 134 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.2.3+2.3.4+...+\text{n(n+1)(n+2)}=\frac{\text{n(n+1)(n+2)(n+3)}}{4}.$
AnswerLet the given satement be P(n), i.e., $\text{P(n)}:1.2.3+2.3.4+...+\text{n(n+1)(n+2)}=\frac{\text{n(n+1)(n+2)(n+3)}}{4}$ For n = 1, we have$\text{P}(1):1.2.3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.2.3.4}{4}=6$
which is true. Let P(k) be true for some positive integer k, i.e.,$1.2.3+2.3.4+...+\text{k(k+1)(k+2)}=\frac{\text{k(k+1)(k+2)(k+3)}}{4}\ \ ...(\text{i})$
We shall now prove that P(k + 1) is true. Consider, 1.2.3 + 2.3.4 + .... + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3) = {1.2.3 + 2.3.4 + ... + k(k + 1)(k + 2)} + (k + 1)(k + 2)(k + 3)$=\frac{\text{k(k+1)(k+2)(k+3)}}{4}+(\text{k+1)(k+2)(k+3)}\ \ [\text{Using (i)}]$
$=(\text{k+1)(k+2)(k+3)}\Big(\frac{\text{k}}{4}+1\Big)$
$=\ \frac{(\text{k+1)(k+2)(k+3)(k+4)}}{4}$
$=\ \frac{(\text{k+1)(k+1+1)(k+1+2)(k+1+3)}}{4}$
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 144 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{n}+1)(2\text{n}+3)}=\frac{\text{n}}{3(2\text{n}+3)}.$
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{n}+1)(2\text{n}+3)}=\frac{\text{n}}{3(2\text{n}+3)}$
For n = 1, we have$\text{P(1)}:\frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{3.5},$
which is true. Let P(k) be true for some positive integer k, i.e.,$\text{P(k)}:\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{k}+1)(2\text{k}+3)}=\frac{\text{k}}{3(2\text{k}+3)}\ ....(1)$
We shall now prove that P(k + 1) is true. Consider$\Big[\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{k}+1)(2\text{k}+3)}\Big]+\frac{1}{\{2(\text{k+1})+1\}\{2(\text{k}+1)+3\}}$
$=\frac{\text{k}}{3(2\text{k}+3)}+\frac{1}{(2\text{k}+3)(2\text{k}+5)}\ \ [\text{Using (1)}]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{\text{k}}{3}+\frac{1}{(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{\text{k}(2\text{k+5})+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{2\text{k}^2+5\text{k}+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{2\text{k}^2+2\text{k}+3\text{k}+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k+3})}\Big[\frac{2\text{k}(\text{k+1})+3(\text{k+1})}{3(2\text{k+5})}\Big]$
$=\frac{(\text{k+1})(2\text{k+3})}{3(2\text{k}+3)(2\text{k+5})}$
$=\frac{(\text{k+1})}{3\{2(\text{k+1})+3\}}$
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 154 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$10^{2\text{n}–1} + 1$ is divisible by 11.
Answer$\text{P(n)}:10^{2\text{n-1}}+1$ is divisible by 11.It can be observed that P(n) is true for n = 1 since
$\text{P(1)}=10^{2.1-1}+1=11,$ which is divisible by 11.
Let P(k) be true for some positive integer k, i.e.,
$10^{2\text{k}-1}+1$ is divisible by 11.
$\therefore\ 10^{2\text{k}-1}+1=11\text{m, where m}\in\text{N}\ ...(1)$
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider,
$=10^{2(\text{k}+1)-1}+1$
$=10^{2\text{k}+2-1}+1$
$=10^{2\text{k}+1}+1$
$=10^2(10^{2\text{k}-1}+1-1)+1$
$=10^2(10^{2\text{k}-1}+1)-10^2+1$
$=10^2.11\text{m}-100+1\ \ [\text{Using (1)}]$
$=100\times11\text{m}-99$
$=11(100\text{m}-9)$
$=11\text{r, where r =}(100\text{m}-9)$ is some natural number
Therefore, $10^{2(\text{k}+1)-1}+1$ is divisible by 11.
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 164 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$3^{2\text{n}+2}-8\text{n}-9$ is divisible by 8.
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:3^{2\text{n}+2}-8\text{n}-9$ is divisible by 8.
It can be observed that P(n) is true for n = 1 since $3^{2\times1+2}-8\times1-9=64,$ which is divisible by 8.Let P(k) be true for some positive integer k, i.e.,
$3^{2\text{k}+2}-8\text{k}-9$ is divisible by 8.
$\therefore\ 3^{2\text{k}+2}-8\text{k}-9=8\text{m};\ \text{where m}\in\text{N}\ ....(1)$
We shall now prove that P(k + 1) is true whenever P(k) is true. Consider,$3^{2(\text{k+1})+2}-8(\text{k+1})-9$
$=3^{2\text{k+2}}.3^2-8\text{k}-8-9$
$=3^2(3^{2\text{k+2}}-8\text{k}-9+8\text{k}+9)-8\text{k}-17$
$=3^2(3^{2\text{k+2}}-8\text{k}-9)+3^2(8\text{k}+9)-8\text{k}-17$
$=9.8\text{m}+9(8\text{k}+9)-8\text{k}-17$
$=9.8\text{m}+72\text{k}+81-8\text{k}-17$
$=9.8\text{m}+64\text{k}+64$
$=8(9\text{m}+8\text{k}+8)$
Therefore, $3^{2(\text{k+1})+2}-8(\text{k+1})-9$ is divisible by 8. 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 174 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\text{x}^{2\text{n}}-\text{y}^{2\text{n}}$ is divisible by $x + y$.
AnswerLet the given statement be P(n), i.e., $P(n): x^{2n} – y^{2n}$ is divisible by $x + y$. It can be observed that P(n) is true for n = 1. This is so because $x^{2 \times 1} – y^{2 \times 1} = x^2 – y^2 = (x + y) (x – y)$ is divisible by (x + y). Let P(k) be true for some positive integer k, i.e., $x^{2k} – y^{2k}$ is divisible by x + y.
$\therefore x^{2k} – y^{2k} = m (x + y)$, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true. Consider,$\text{x}^{2(\text{k+1})}-\text{y}^{2(\text{k}+1)}$
$=\text{x}^{2\text{k}}.\text{x}^2-\text{y}^{2\text{k}}.\text{y}^2$
$=\text{x}^2(\text{x}^{2\text{k}}-\text{y}^{2\text{k}}+\text{y}^{2\text{k}})-\text{y}^{2\text{k}}.\text{y}^2$
$=\text{x}^2\{\text{m}(\text{x+y)}+\text{y}^{2\text{k}}\}-\text{y}^{2\text{k}}.\text{y}^2\ \ [\text{Using (1)}]$
$=\text{m(x+y)x}^2+\text{y}^{2\text{k}}.\text{x}^2-\text{y}^{2\text{k}}.\text{y}^2$
$=\text{m(x+y)x}^2+\text{y}^{2\text{k}}(\text{x}^2-\text{y}^2)$
$=\text{m(x+y)x}^2+\text{y}^{2\text{k}}(\text{x}+\text{y})(\text{x}-\text{y})$
$=(\text{x+y})\{\text{mx}^2+\text{y}^{2\text{k}}(\text{x}-\text{y})\},$ which is a factor of (x + y).
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 184 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1^3+2^3+3^3+....+\text{n}^3=\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2.$
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:1^3+2^3+3^3+....+\text{n}^3=\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
For n = 1, we have$\text{P}(1):1^3=1=\Big(\frac{1(1+1)}{2}\Big)^2=\Big(\frac{1.2}{2}\Big)^2=1^2=1$
which is true. Let P(k) be true for some positive integer k, i.e.,$1^3+2^3+3^3+....+\text{k}^3=\Big(\frac{\text{k(k+1)}}{2}\Big)^2\ \ ....(\text{i})$
We shall now prove that P(k + 1) is true. Consider,$1^3+2^3+3^3+...+\text{k}^3+(\text{k}+1)^3$
$=(1^3+2^3+3^3+...+\text{k}^3)+(\text{k}+1)^3$
$=\Big(\frac{\text{k(k+1)}}{2}\Big)^2+(\text{k}+1)^3\ \ [\text{Using (i)}]$
$=\frac{\text{k}^2(\text{k+1})^2}{4}+(\text{k}+1)^3$
$=\frac{\text{k}^2(\text{k+1})^2+4(\text{k}+1)^3}{4}$
$=\frac{(\text{k+1})^2\{\text{k}^2+4(\text{k+1})\}}{4}$
$=\frac{(\text{k+1})^2\{\text{k}^2+4\text{k+4}\}}{4}$
$=\frac{(\text{k}+1)^2(\text{k}+2)^2}{4}$
$=\frac{(\text{k}+1)^2(\text{k}+1+1)^2}{4}$
$=\Big(\frac{(\text{k+1})(\text{k+1+1})}{2}\Big)^2$
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 number i.e., n.
View full question & answer→Question 194 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.2+2.3+3.4+...+\text{n.(n+1)}=\Big[\frac{\text{n(n+1)(n+2)}}{3}\Big].$
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:1.2+2.3+3.4+...+\text{n.(n+1)}=\Big[\frac{\text{n(n+1)(n+2)}}{3}\Big]$
For n = 1, we have$\text{P(1)}:1.2=2=\frac{1(1+1)(1+2)}{3}=\frac{1.2.3}{3}=2,$
which is true. Let P(k) be true for some positive integer k, i.e.,$1.2+2.3+3.4+...+\text{k}.(\text{k+1})=\Big[\frac{\text{k(k+1)(k+2)}}{3}\Big]\ \ ....(\text{i})$
We shall now prove that P(k + 1) is true. Consider,$1.2+2.3+3.4+....+\text{k.(k+1)+(k+1).(k+2)}$
$=[1.2+2.3+3.4+....+\text{k.(k+1)]+(k+1).(k+2)}$
$=\frac{\text{k(k+1)(k+2)}}{3}+(\text{k+1)(k+2)}\ \ [\text{Using (i)}]$
$=(\text{k+1)(k+2)}\Big(\frac{\text{k}}{3}+1\Big)$
$=\frac{(\text{k+1)(k+2)(k+3)}}{3}$
$=\frac{(\text{k+1)(k+1+1)(k+1+2)}}{3}$
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 204 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.2+2.2^2+3.2^3+...+\text{n}.2^\text{n}=(\text{n}-1)2^{\text{n+1}}+2.$
AnswerLet the given statement be P(n), i.e., $\text{P(n)}:1.2+2.2^2+3.2^3+...+\text{n}.2^\text{n}=(\text{n}-1)2^{\text{n+1}}+2$ For n = 1, we have$\text{P(1)}:1.2=2=(1-1)2^{1+1}+2=0+2=2,$
which is true. Let P(k) be true for some positive integer k, i.e., $1.2+2.2^2+3.2^3+...+\text{k}.2^\text{k}=(\text{k}-1)2^{\text{k}+1}+2\ \ ....(\text{i})$ We shall now prove that P(k + 1) is true. Consider,$\{1.2+2.2^2+3.2^3+...+\text{k}.2^{\text{k}}\}+(\text{k}+1).2^{\text{k}+1}$
$=(\text{k}-1)2^{\text{k}+1}+2+(\text{k}+1)2^{\text{k+1}}$
$=2^{\text{k+1}}\{(\text{k}-1)+(\text{k}+1)\}+2$
$=2^{\text{k}+1}.2\text{k}+2$
$=\text{k}.2^{(\text{k}+1)+1}+2$
$=\{(\text{k+1})-1\}2^{(\text{k}+1)+1}+2$
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 214 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.3+3.5+5.7+...+(2\text{n}-1)(2\text{n}+1)=\frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}.$
Answer$\text{Let P(n)}=1.3+3.5+5.7+...+(2\text{n}-1)(2\text{n}+1)=\frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$For n = 1
$\text{P}(1)=(2\times1-1)(2\times1+1)=\frac{1[4(1)^2+6\times1-1]}{3}$
⇒ 3 = 3
$\therefore$ P(1) is true.
Now, let P(n) be true for n = k
$\therefore\ \text{P(k)}=1.3+3.5+5.7+...+(2\text{k}-1)(2\text{k}+1)=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}$
For n = k + 1
$\text{P(k+1)}=1.3+3.5+5.7+...+(2\text{k}-1)(2\text{k}+1)+[2(\text{k}+1)-1][2(\text{k}+1)+1]$
$=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}+[2(\text{k+1})-1][2(\text{k+1})+1]$
$\Rightarrow\ \text{P(k+1)}=\frac{4\text{k}^3+6\text{k}^2-\text{k}}{3}+(2\text{k}+1)(2\text{k}+3)$
$=\frac{4\text{k}^3+6\text{k}^2-\text{k}+3(4\text{k}^2+8\text{k}+3)}{3}$
$\Rightarrow\ \text{P(k+1)}=\frac{4\text{k}^3+6\text{k}^2-\text{k}+12\text{k}^2+24\text{k}+9}{3}$
$=\frac{4\text{k}^3+18\text{k}^2+23\text{k}+9}{3}$
$=\frac{(\text{k+1})(4\text{k}^2+14\text{k}+9)}{3}$
$\therefore$ 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 224 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...\text{n})}=\frac{2\text{n}}{(\text{n+1})}.$
Answer$\text{Let}\ \text{P(n)}=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+\text{n})}=\frac{2\text{n}}{(\text{n+1})}$For n = 1,
$\text{P(1)}=1=\frac{2\times1}{1+1}$
⇒ 1 = 1
$\therefore$ P(1) is true.
Now, let P(n) be true for n = k
$\therefore\ \text{P(k)}=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+\text{k})}=\frac{2\text{k}}{(\text{k+1})}\ ....(\text{i})$
For n = k + 1,
$\text{P(k+1)}=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+\text{k})}$
$+\frac{1}{(1+2+3+...+\text{k+1})}=\frac{2\text{k}}{(\text{k+1})}+\frac{1}{(1+2+3+...+\text{k+1})}\ \ [\text{Using (i)}]$
$\Rightarrow\ \text{P(k+1)}=\frac{2\text{k}}{\text{k}+1}+\frac{1}{\frac{(\text{k+1})(\text{k+2})}{2}}$
$=\frac{2\text{k}}{\text{k+1}}+\frac{2}{(\text{k+1})(\text{k+2})}$
$=\frac{2}{\text{k+1}}\Big[\frac{(\text{k+1})^2}{\text{k}+2}\Big]=\frac{2(\text{k+1})}{\text{k+2}}$
$\therefore$ 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 234 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\text{n(n+1)(n+2)}}=\frac{\text{n(n+3)}}{4(\text{n+1})(\text{n}+2)}.$
Answer$\text{Let }\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\text{n(n+1)(n+2)}}=\frac{\text{n(n+3)}}{4(\text{n+1})(\text{n}+2)}$For n = 1
$\text{P(1)}=\frac{1}{1(1+1)(1+2)}=\frac{1(1+3)}{4(1+1)(1+2)}$
$\Rightarrow\ \frac{1}{6}=\frac{1}{6}$
$\therefore$ P(1) is true.
Now, let P(n) be true for n = k
$\therefore\ \text{P(k)}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\text{k(k+1)(k+2)}}=\frac{\text{k(k+3)}}{4(\text{k+1})(\text{k}+2)}\ \ ...(\text{i})$
For n = k + 1
$\Rightarrow\ \text{R.H.S.}=\frac{(\text{k+1})(\text{k+4})}{4(\text{k+2})(\text{k+3})}$
And $\text{L.H.S.}=\frac{\text{k(k+3)}}{4(\text{k+2})(\text{k}+3)}+\frac{1}{(\text{k+1})(\text{k+2})(\text{k+3})}\ \ [\text{Using (i)}]$
$=\frac{1}{(\text{k+1})(\text{k+2})}\Big[\frac{\text{k}^2+3\text{k}}{4}+\frac{1}{\text{k}+3}\Big]$
$=\frac{1}{(\text{k+1})(\text{k+2})}\Big[\frac{\text{k}^3+6\text{k}^2+9\text{k}+4}{4(\text{k+3})}\Big]$
$=\frac{1}{(\text{k+1})(\text{k+2})}\Big[\frac{(\text{k+1)}^2(\text{k+4)}}{4(\text{k+3})}\Big]$
$=\frac{(\text{k+1})(\text{k+4})}{4(\text{k+2})(\text{k}+3)}$
$\therefore$ 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 $\text{n}\in\text{N}.$
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