Question

Найти сумму:

1/(1*2*3)+1/(2*3*4)+...+1/(n(n+1)(n+2))

Answer 1

$frac{1}{1*2*3}+frac{1}{2*3*4}+...+frac{1}{n(n+1)(n+2)}= frac {1}{2} *(frac{3-1}{1*2*3}+frac{4-2}{2*3*4}+...+frac{(n+2)-n}{n(n+1)(n+2)})= frac{1}{2}*(frac{1}{1*2}-frac{1}{2*3}+frac{1}{2*3}-frac{1}{3*4}+...+frac{1}{n(n+1)}-frac{1}{(n+1)(n+2)})= frac{1}{2}*(frac{1}{1*2}-frac{1}{(n+1)(n+2)})= frac{1}{2}*(frac{(n+1)(n+2)-1*2}{1*2(n+1)(n+2)})=frac{n^2+3n+2-2}{4(n+1)(n+2)}=frac{n^2+3n}{4(n+1)(n+2)}=frac{n(n+3)}{4(n+1)(n+2)}$