Question

Prove $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{n(n+1)(n+2)}=\frac{1(n+3)}{4(n+1)(n+2)}$

Answer 1

p(1) = $\frac{1}{6}=\frac{1\cdot 4}{4(2\cdot 4)}=\frac{1}{5}$
Similary p (2) also holds
Assums p(n) also holds
$\therefore p(n+1)=p\left(n\right)+\frac{1}{(n+1)(n+2)(n+3)}$
$=\frac{n(n+3)}{4(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}$
$=\frac{1}{4(n+1)(n+2)(n+3)}\left[n\right(n+3{)}^2+4]$
$\frac{n^3+6n^2+9n+4}{4(n+1)(n+2)(n+3)}$
$=\frac{(n+1{)}^2(n+4)}{4(n+1)(n+2)(n+3)}$
$=\frac{(n+1)(n+4)}{4(n+1)(n+2)}$
$=\frac{N(N+3)}{4(N+1)(M+2)}$
Where N = n + 1