(Each question 4 marks) - MATHS STD 11 Science Questions - Vidyadip

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Prove the following by the principle of mathematical induction: $\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$

Answer

Let 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

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Question 24 Marks

Prove the following by the principle of mathematical induction: $1 + 3 + 3^2 + ... + 3^{\text{n}-1}=\frac{3^\text{n}-1}{2}$

Answer

Let 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 $\text{n}\in\text{N}.$

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Question 34 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 $\text{n}\in\text{N}.$

Answer

Let 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.

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Question 44 Marks

Prove the following by the principle of mathematical induction: $1.3 + 3.5 + 5.7 + ... + (2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$

Answer

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 $1.3=\frac{1(4+6-1)}{3}$ 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

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Question 54 Marks

Prove that $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$ for all $\text{n}\in\text{N}.$

Answer

P(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

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Question 64 Marks

Prove the following by the principle of mathematical induction: 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! - 1 for all $\text{n}\in\text{N}$

Answer

Consider 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

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Question 74 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 $\text{n}\in\text{N}.$

Answer

Let $p(n)$ be the statement given by $P(n): a_n=3 \times 7^{n-1}$ for all $\mathbf{n} \in \mathbf{N}$.
Step 1: $p(2): a_2=3 \times 7^{2-1}=21$ Given that $a_k=7 a_{k-1}$ for all natural numbers $k \geq 2 a_2=7 a_1=7 \times 3=21$
$\therefore p(2)$ 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.
$a_{m+1}=7 a_m a_{m+1}=7 \times a_m a_{m+1}=7^1 \times 3 \times 7^{m-1} \ldots$ [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 $\mathbf{n} \in \mathbf{N}$.

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Question 84 Marks

Prove the following by the principle of mathematical induction: $5^{2 n+2}+24 n-25$ is divisible by 576 for all $\mathbf{n} \in \mathbf{N}$

Answer

Let $p(n): 5^{2 n+2}+24 n-25$ is divisible by 576 for all 576 For $n=15^4-24-25=625-49=576$ Which is divisible by $576 \Rightarrow p(n)$ is true for $n=1$ Let $p(n)$ is true for $n=k$, so $5^{2 k+2}-24 k-25$ is divisible by 576
$5^{2 \mathbf{k}+2}-24 \mathbf{k}-25=576 \lambda \ldots$ (1) We have to show that, $5^{(2 k+2)+2}-24(k+1)-25$ is divisible by 576
$5^{(2 \mathbf{k}+2)+2}-24(\mathbf{k}+1)-25=576 \mu$ Now, $5^{(2 \mathbf{k}+2)+2}-24(\mathbf{k}+1)-25=5^{(2 \mathbf{k}+2)} .5^2-24 \mathbf{k}-24-25$
$(576 \lambda+24 \mathrm{k}+25) 25-24 \mathrm{k}-49$ [Using equation (1)] $=25.576 \lambda+600 \mathrm{k}+625-24 \mathrm{k}-49$
$=25.576 \lambda+576 \mathrm{k}+576=576(25 \lambda+\mathrm{k}+1)=576 \mu \Rightarrow \mathrm{p}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{k}+1 \Rightarrow \mathrm{p}(\mathrm{n})$ is true for all by $\mathbf{n} \in \mathbf{N P M I}$

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Question 94 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 $n^2$.

Answer

Let $P(n): 1+3+5+\ldots+(2 n-1)=n^2$ 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 $P(k): 1+3+5+\ldots+(2 k-1)=k^2 \ldots$ (1)
We have to show that $1+3+5+\ldots+(2 k-1)+2(k+1)-1=(k+1)^2$
Now, $\{1$ $+3+5+\ldots+(2 k-1)\}+2(k+1)=k^2+2(k+1)\left[\right.$ Using equation (1)] $=k^2+2 k+1=(k+1)^2 \Rightarrow P(n)$ is true for $n=k$ $+1 \Rightarrow P(n)$ is true for all $\mathbf{n} \in N$ by PMI

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Question 104 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: $\text{S}_\text{k+1}=\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}+2}$ $=\frac{-1}{\text{k}+1}+\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$ $=\frac{-1}{\text{k}+1}+\text{S}_\text{k}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$ $=\text{S}_\text{k}+\frac{1}{2(2\text{k}+1)(\text{k}+1)}$ $>\text{S}_\text{k}$ $\therefore\text{S}_\text{k+1}>\frac{13}{24}$

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Question 114 Marks

Prove the following by the principle of mathematical induction: $3^{2n} + 7$ is divisible by $8$ for all $\text{n}\in\text{N}$

Answer

Let $p(n): 3^{2 n}+7$ is divisible by 8 for all $\mathbf{n} \in N$ For $n=13^2+7=16$ Which is divisible by $8 \Rightarrow p(n)$ is true for $n=1$ Let $p(n)$ is true for $n=k$, so $3^{2 k}+7$ is divisible by $8 \Rightarrow 3^{2 k}+7=8 \lambda \ldots$ (1)
We have to show that, $3^{2(k+1)}+7$ is divisible by $8\ 3^{(2 \mathbf{k}+1)}+7=8 \mu$
Now, $3^{2(k+1)}+7=3^{2 k} \cdot 3^2+7=9.3^{2 k}+7=9 .(8 \lambda-7)+7=72 \lambda-56$
$=8(9 \lambda-7)=8 \mu \Rightarrow p(n)$ is true for $n=k+1 \Rightarrow p(n)$ is true for all by $\mathbf{n} \in N$ PMI

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Question 124 Marks

Prove the following by the principle of mathematical induction: $1.3 + 2.4 + 3.5 + ... + \text{n}(\text{n} + 2) $ $=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+7)$

Answer

Let $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

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Question 134 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 $=\frac{\text{n}+1}{2\text{n}}$

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Question 144 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 $\frac{\text{n}(\text{n}+1)}{2}.$

Answer

Let 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 $\text{n}\in\text{N}$

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Question 154 Marks

Prove the following by the principle of mathematical induction: $11^{n+2}+12^{2 n+1}$ is divisible of $133$ for all $\text{n}\in\text{N}$

Answer

Let $p(n): 11^{n+2}+12^{2 n+1}$ is divisible of 133 For $n=111^3+12^3=1331+1728=3059$ 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 $13311^{k+2}+12^{2 k+1}=133 \lambda \ldots$ (1)
We have to show that, $1^{k+3}+12^{2 k+3}$ is divisible of 133
Now, $11^{k+2} \cdot 11+12^{2 k+1} \cdot 12^2$ $=\left(133 \lambda-12^{2 \mathbf{k}+1}\right) 11+12^{2 \mathbf{k}+1} .144=11.133 \lambda+11.12^{2 \mathbf{k}+1}+144.12^{2 \mathbf{k}+1}=11.133 \lambda+133.12^{2 \mathbf{k}+1}$ $=133\left(11 \lambda+12^{2 \mathrm{k}+1}\right)=133 \mu \Rightarrow \mathrm{p}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{k}+1 \Rightarrow \mathrm{p}(\mathrm{n})$ is true for all by $\mathbf{n} \in \mathrm{N}$ PMI

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Question 164 Marks

Prove the following by the principle of mathematical induction: $1^2 + 2^2 + 3^2 +...+ \text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$

Answer

Let 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 $\text{P}(1):1=\frac{1(1+1)(2+1)}{6}$
$1=1 \Rightarrow$ 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

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Question 174 Marks

Prove that the number of subsets of a set containing $\mathbf{n}$ distinct elements is $2^n$ for all $\mathbf{n} \in \mathbf{N}$.

Answer

Let $p(n)$ be the statement given by $P(n)$ : The number of subsets of a set containing $n$ distinct element is is $2^n$ for all $\mathbf{n} \in \mathbf{N}$. 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 ... (1) 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 $\mathbf{n} \in \mathbf{N}$.

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Question 184 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.

Answer

Let 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.

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Question 194 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 $\text{n}\in\text{N}$

Answer

Let 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 $\text{n}\in\text{N}.$

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Question 204 Marks

Prove the following by the principle of mathematical induction: $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{n(n+1)}}=\frac{\text{n}}{\text{n}+1}$

Answer

Let 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

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Question 214 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 $\text{n}\in\text{N}.$

Answer

Let 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

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Question 224 Marks

Prove the following by the principle of mathematical induction: $2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$ $$

Answer

Let 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.

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