Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}$

(N/A) Let the given statement be $P(n)$:
$P(n): \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}$
For $n=1$,we have:
$P(1): \frac{1}{1 \cdot 4} = \frac{1}{4}$ and $\frac{1}{3(1)+1} = \frac{1}{4}$.
Since $\frac{1}{4} = \frac{1}{4}$,$P(1)$ is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$P(k): \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \ldots + \frac{1}{(3k-2)(3k+1)} = \frac{k}{3k+1}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left\{ \frac{1}{1 \cdot 4} + \ldots + \frac{1}{(3k-2)(3k+1)} \right\} + \frac{1}{\{3(k+1)-2\}\{3(k+1)+1\}}$
$= \frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)}$ [Using $(i)$]
$= \frac{1}{3k+1} \left\{ k + \frac{1}{3k+4} \right\}$
$= \frac{1}{3k+1} \left\{ \frac{k(3k+4) + 1}{3k+4} \right\}$
$= \frac{3k^2 + 4k + 1}{(3k+1)(3k+4)}$
$= \frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$
$= \frac{k+1}{3k+4} = \frac{k+1}{3(k+1)+1}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.