# Prove the following by using the principle of mathematical induction for all n ∈ N:

Question:

Prove the following by using the principle of mathematical induction for all n ∈ N$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}$

Solution:

Let the given statement be $\mathrm{P}(n)$, i.e.,

$\mathrm{P}(n): \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}$

For n = 1, we have

$P(1): \frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{3.5}$, which is true.

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$\mathrm{P}(k): \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}=\frac{k}{3(2 k+3)}$ (1)

We shall now prove that P(k + 1) is true.

Consider

$\left[\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}\right]+\frac{1}{\{2(k+1)+1\}\{2(k+1)+3\}}$

$=\frac{k}{3(2 k+3)}+\frac{1}{(2 k+3)(2 k+5)}$  $[$ Using (1) $]$

$=\frac{1}{(2 k+3)}\left[\frac{k}{3}+\frac{1}{(2 k+5)}\right]$

$=\frac{1}{(2 k+3)}\left[\frac{k(2 k+5)+3}{3(2 k+5)}\right]$

$=\frac{1}{(2 k+3)}\left[\frac{2 k^{2}+5 k+3}{3(2 k+5)}\right]$

$=\frac{1}{(2 k+3)}\left[\frac{2 k^{2}+2 k+3 k+3}{3(2 k+5)}\right]$

$=\frac{1}{(2 k+3)}\left[\frac{2 k(k+1)+3(k+1)}{3(2 k+5)}\right]$

$=\frac{(k+1)(2 k+3)}{3(2 k+3)(2 k+5)}$

$=\frac{(k+1)}{3\{2(k+1)+3\}}$

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.