Question

Q23. Prove by mathematical induction that...​

Answer 1

Definition :-

The Principle of Mathematical Induction

• Suppose, there is a given statement P(n) where n is a natural number, such that

• 1. P(n) is true for n = 1.

• 2. If the statement P(n) is true for n = k, where k is natural number, then the statement P(n) is true for n = k + 1.

• Then, this implies, P(n) is true for all natural numbers.

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: P(n) : \frac{1}{3.5} + \frac{1}{5.7} + - - - - + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}$

Step :- 1

• For n = 1,

we have

$\rm :\longmapsto\:\dfrac{1}{3.5} = \dfrac{1}{3(2 \times 1 + 3)}$

$\rm :\longmapsto\:\dfrac{1}{15} = \dfrac{1}{15}$

$\rm :\implies\:P(n) \: is \: true \: for \: n \: = \: 1$

Step :- 2

• Let suppose that P(n) is true for n = k.

Therefore,

$\sf \:\frac{1}{3.5} + \frac{1}{5.7} + - - - - + \frac{1}{(2k + 1)(2k + 3)} = \frac{k}{3(2k + 3)} - - - (1)$

Step :- 3

For n = k + 1,

• we have to prove that P(n) is true.

$\sf \:\frac{1}{3.5} + \frac{1}{5.7} + - - + \frac{1}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 3)(2k + 5)} = \frac{k + 1}{3(2k + 5)}$

Consider, LHS

$\sf \:\dfrac{1}{3.5} + \dfrac{1}{5.7} + - - + \dfrac{1}{(2k + 1)(2k + 3)} + \dfrac{1}{(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{k}{3(2k + 3)} + \dfrac{1}{(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{k(2k + 5) + 3}{3(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{ {2k}^{2} + 5k + 3}{3(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{ {2k}^{2} + 3k + 2k + 3}{3(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{k(2k + 3) + 1(2k + 3)}{3(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{(2k + 3)(k + 1)}{3(2k + 3)(2k + 5)}$

$\sf \: = \: \: \dfrac{(k + 1)}{3(2k + 5)}$

$\rm :\implies\:P(n) \: is \: true \: for \: n \: = \: k + 1$

Hence,

• By the Principle of Mathematical Induction,

$\sf \:\dfrac{1}{3.5} + \dfrac{1}{5.7} + - - + \dfrac{1}{(2n + 1)(2n + 3)} = \dfrac{n}{3(2n + 3)}$

Answer 2

ANSWER:

To Prove:

$\:\:\:\:\bullet\:\:\:\:P(n)=\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\cdots\cdots+\dfrac{1}{(2n+1)\times(2n+3)}=\dfrac{n}{3(2n+3)}\\\\\:\:\:\:\:\:\:\:\text{(By Principle of Mathematical Induction)}$

Proof:

$:\longrightarrow P(n)=\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\cdots\cdots+\dfrac{1}{(2n+1)\times(2n+3)}=\dfrac{n}{3(2n+3)}\\\\\text{\underline{STEP 1: Put n = 1}}\\\\\text{LHS}:\implies\dfrac{1}{3\times5}=\dfrac{1}{15}- - - -(i)\\\\\text{RHS}:\implies\dfrac{1}{3(2(1)+3)}=\dfrac{1}{3(5)}=\dfrac{1}{15}- - - -(ii)\\\\\text{Hence, P(1) is true.}\\\\\text{\underline{STEP 2 : Let, P(k) be true.}}\\\\:\implies P(k)=\dfrac{1}{3(2k+3)}\\\\\text{\underline{STEP 3 : To Prove-}}$

$:\implies P(k+1)=\dfrac{k+1}{3(2(k+1)+3)}=\dfrac{k+1}{3(2k+5)}\\\\\text{\underline{STEP 4 : Proof- }}\\\\\text{Taking LHS,}\\\\:\implies P(k+1)\\\\:\implies\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\cdots\cdots+\dfrac{1}{(2(k+1)+1)\times(2(k+1)+3)}\\\\:\implies\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\cdots\cdots+\dfrac{1}{(2k+1)\times(2k+3)}+\dfrac{1}{(2(k+1)+1)\times(2(k+1)+3)}$

$:\implies\underbrace{\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\cdots\cdots+\dfrac{1}{(2k+1)\times(2k+3)}}_{\text{P(k)}}+\dfrac{1}{(2k+3)\times(2k+5)}\\\\:\implies P(k)+\dfrac{1}{(2k+3)\times(2k+5)}\\\\:\implies\dfrac{k}{3(2k+3)}+\dfrac{1}{(2k+3)\times(2k+5)}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{k(2k+5)+3}{(3)(2k+3)(2k+5)}\\\\:\implies\dfrac{2k^2+5k+3}{(3)(2k+3)(2k+5)}\\\\:\implies\dfrac{2k^2+2k+3k+3}{(3)(2k+3)(2k+5)}\\\\:\implies\dfrac{2k(k+1)+3(k+1)}{(3)(2k+3)(2k+5)}$

$:\implies\dfrac{(2k+3)(k+1)}{(3)(2k+3)(2k+5)}\\\\\text{(2k+3) gets cancelled. So,}\\\\:\implies\dfrac{k+1}{3(2k+5)} = \text{RHS}\\\\\text{ \because LHS = RHS}\\\\\text{ \therefore P(k+1) is true, when P(k) is true.}\\\\\text{\bf{Hence, P(n) is True!}}$