Question:

$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.5}+\ldots+\frac{1}{(4 n-1)(4 n+3)}=\frac{n}{3(4 n+3)}$

Solution:

LetÂ P(n) be the given statement.

Now,

$P(n)=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\ldots+\frac{1}{(4 n-1)(4 n+3)}=\frac{n}{3(4 n+3)}$

Step 1:

$P(1)=\frac{1}{3.7}=\frac{1}{21}=\frac{1}{3(4+3)}$

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ is true.

Then,

$\frac{1}{3.7}+\frac{1}{7.11}+\ldots+\frac{1}{(4 m-1)(4 m+3)}=\frac{m}{3(4 m+3)}$

To prove: $P(m+1)$ is true.

That is,

$\frac{1}{3.7}+\frac{1}{7.11}+\ldots+\frac{1}{(4 m+3)(4 m+7)}=\frac{m+1}{3(4 m+7)}$

Now,

$P(m)=\frac{1}{3.7}+\frac{1}{7.11}+\ldots+\frac{1}{(4 m-1)(4 m+3)}=\frac{m}{3(4 m+3)}$

$\Rightarrow \frac{1}{3.7}+\frac{1}{7.11}+\ldots+\frac{1}{(4 m-1)(4 m+3)}+\frac{1}{(4 m+3)(4 m+7)}=\frac{m}{3(4 m+3)}+\frac{1}{(4 m+3)(4 m+7)}$

$\left[\right.$ Adding $\frac{1}{(4 m+3)(4 m+7)}$ to both sides $]$

$\Rightarrow \frac{1}{3.7}+\frac{1}{7.11}+\ldots+\frac{1}{(4 m+3)(4 m+7)}=\frac{4 m^{2}+7 m+3}{3(4 m+3)(4 m+7)}=\frac{(4 m+3)(m+1)}{3(4 m+3)(4 m+7)}=\frac{m+1}{3(4 m+7)}$

Thus, $P(m+1)$ is true.

By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.