Question

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

Answer 1

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots +\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$
$\text{For n = 1}P\left(1\right)=\frac{1}{(3\times 1-2)(3\times 1+1)}=\frac{1}{(3\times 1+1)}\Rightarrow \frac{1}{1\times 4}=\frac{1}{4}\Rightarrow \frac{1}{4}=\frac{1}{4}\therefore \text{P (1) is true}\text{Let P (n) be true for n = k}\therefore P\left(k\right)=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots +\frac{1}{(3n-2)(3k+1)}=\frac{n}{(3k+1)}\text{For n = k + 1}\text{R.H.S.}=\frac{k+1}{(3k+1)}\text{L.H.S.}=\frac{k}{(3k+1)}+\frac{1}{(3k+1)(3k+4)}=\frac{1}{(3k+1)}[k+\frac{1}{3k+4}]=\frac{1}{(3k+1)}\left[\frac{3k^2+4k+1}{3k+4}\right]=\frac{1}{(3k+1)}\left[\frac{3k62+3k+k+1}{3k+4}\right]=\frac{1}{(3k+1)}\left[\frac{(3k+1)(k+1)}{3k+4}\right]=\frac{k+1}{3k+4}$
$\therefore$ P(k + 1) is true
thus P(k) is true $\Rightarrow P(k+1)$ is true
Hence by principle fo mathematical induction,