Question

Вычислить сумму:

$\frac{1}{1*5} + \frac{1}{5*9} +... + \frac{1}{(4n-3)(4n+1)}$

И пожалуйста с объяснениями

Answer 1

$\frac{1}{(4n-3)(4n+1)}=\frac{1/4}{4n-3}-\frac{1/4}{4n+1}=\frac{1}{4}\cdot (\frac{1}{4n-3}-\frac{1}{4n+1})\; \; \; \Rightarrow \\\\\\\frac{1}{1\cdot 5}+\frac{1}{5\cdot 9}+...+\frac{1}{(4n-3)(4n+1)}=\\\\=\frac{1}{4}\cdot \Big ((\frac{1}{1}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{13})+...+(\frac{1}{4n-7}-\frac{1}{4n-3})+(\frac{1}{4n-3}-\frac{1}{4n+1})\Big )=\\\\=\frac{1}{4}\cdot (1-\frac{1}{4n+1})=\frac{1}{4}\cdot \frac{4n}{4n+1}=\frac{n}{4n+1}$