Question

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{6n+4}$

Answer 1

Let P (n) $=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{6n+4}$
For n = 1
$P\left(1\right)=\frac{1}{(3\times 1-1)(3\times 1+2)}=\frac{1}{(6\times 1+4)}\Rightarrow \frac{1}{2\times 5}=\frac{1}{10}\Rightarrow \frac{1}{10}=\frac{1}{10}$
$\therefore$ P (1) true
Let P (n) be true for n = k
$\therefore P\left(k\right)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3k-1)(3k+2)}=\frac{k}{6k+4}Forn=k+1P(k+1)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3k-1)(3k+2)}+\frac{1}{\left[k\right(k+1)-1]\left[3\right(k+1)+2]}=\frac{k+1}{6k+10}=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}=\frac{1}{(3k+2)}[\frac{k}{2}+\frac{1}{3k+5}]=\frac{1}{(3k+2)}\left[\frac{3k^2+5k+2}{2(3k+5)}\right]=\frac{1}{(3k+2)}\left[\frac{3k^2+3k+2k+2}{2(3k+5)}\right]=\frac{1}{(3k+2)}\left[\frac{3k(k+1)+2(k+1)}{2(3k+5)}\right]=\frac{1}{(3k+2)}\left[\frac{(k+1)(3k+2)}{2(3k+5)}\right]=\frac{k+1}{6k+10}$
$\therefore$ P(3k + 10) is true
thus P(k) is true $\Rightarrow$ P(k + 1) is true
Hence by principle fo mathematical induction,
P(n) is true for all$nϵN$