Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$

Let the given statement be $P(n)$,i.e.,
$P(n): \frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$
For $n=1$,we have
$P(1): \frac{1}{2 \times 5} = \frac{1}{10} = \frac{1}{6(1)+4} = \frac{1}{10}$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$P(k): \frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \ldots + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4}$ ..........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\frac{1}{2 \times 5} + \ldots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{\{3(k+1)-1\}\{3(k+1)+2\}}$
$= \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)}$ [Using $(i)$]
$= \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)}$
$= \frac{1}{3k+2} \left( \frac{k}{2} + \frac{1}{3k+5} \right)$
$= \frac{1}{3k+2} \left( \frac{k(3k+5) + 2}{2(3k+5)} \right)$
$= \frac{1}{3k+2} \left( \frac{3k^2 + 5k + 2}{2(3k+5)} \right)$
$= \frac{1}{3k+2} \left( \frac{(3k+2)(k+1)}{2(3k+5)} \right)$
$= \frac{k+1}{2(3k+5)} = \frac{k+1}{6k+10}$
$= \frac{k+1}{6(k+1)+4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.