Question

1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+1/(a+3)(a+4)

Answer 1

Известно, и это можно проверить, сложив дроби,что

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\; \; \; \Rightarrow \\\\\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\frac{1}{(a+2)(a+3)}+\frac{1}{(a+3)(a+4)}=\\\\=(\frac{1}{a}-\frac{1}{a+1})+(\frac{1}{a+1}-\frac{1}{a+2})+(\frac{1}{a+2}-\frac{1}{a+3})+(\frac{1}{a+3}-\frac{1}{a+4})=\\\\=\frac{1}{a}-\frac{1}{a+4}=\frac{a+4-a}{a(a+4)}=\frac{4}{a(a+4)}$