Question:

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}$

Solution:

Let P(n) be the given statement.

Now,

$P(n)=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}$

Step 1;

$P(1)=\frac{1}{1.4}=\frac{1}{4}=\frac{1}{3 \times 1+1}$

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ be true.

i.e.,

$\frac{1}{1.4}+\frac{1}{4.7}+\ldots+\frac{1}{(3 m-2)(3 m+1)}=\frac{m}{3 m+1}$

Now,

$P(m)=\frac{1}{1.4}+\frac{1}{4.7}+\ldots+\frac{1}{(3 m-2)(3 m+1)}=\frac{m}{3 m+1}$

$\Rightarrow \frac{1}{1.4}+\frac{1}{4.7}+\ldots+\frac{1}{(3 m-2)(3 m+1)}+\frac{1}{(3 m+1)(3 m+4)}=\frac{m}{3 m+1}+\frac{1}{(3 m+1)(3 m+4)}$

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

$\Rightarrow \frac{1}{1.4}+\frac{1}{4.7}+\ldots+\frac{1}{(3 m+1)(3 m+4)}=\frac{3 m^{2}+4 m+1}{(3 m+1)(3 m+4)}=\frac{(3 m+1)(m+1)}{(3 m+1)(3 m+4)}=\frac{m+1}{3 m+4}$

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

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in \mathrm{N}$.