Question

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

Answer 1

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

Put n = 1

P(1) : $\frac{1}{1.4}=\frac{1}{4}$

$\frac{1}{4}=\frac{1}{4}$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

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

We have to show that,

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{(3k-2)(3k+1)}+\frac{1}{(3k+1)(3k+4)}=\frac{(k+1)}{(3k+4)}$

Now,

$\{\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......+\frac{1}{(3k-2)(3k+1)}\}+\frac{1}{(3k+1)(3k+4)}$

$=\frac{k}{(3k+1)}+\frac{1}{(3k+1)(3k+4)}$

$=\frac{1}{(3k+1)}[\frac{k}{1}+\frac{1}{(3k+4)}]$

$=\frac{1}{(3k+1)}\left[\frac{k(3k+4)+1}{3k+4}\right]$

$=\frac{1}{(3k+1)}\left[\frac{3k^2+4k+1}{(3k+4)}\right]$

$=\frac{1}{(3k+1)}\left[\frac{3k^2+4k+1}{3k+4}\right]$

$=\frac{1}{(3k+1)}\left[\frac{3k^2+3k+k+1}{3k+4}\right]$

$=\left[\frac{3k(k+1)+(k+1)}{(3k+1)(3k+4)}\right]$

$=\frac{(k+1)(3k+1)}{(3k+1)(3k+4)}$

$=\frac{(k+1)}{(3k+4)}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true all n $ϵ$ N by PMI.