Суммаро ҳисоб кунед:

\(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{(4n-3)\cdot(4n+1)}\)

\(S = \frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{(4n-3)\cdot(4n+1)}\)

\(\frac{1}{1\cdot5}=\frac{1}{4}\cdot(\frac{1}{1}-\frac{1}{5})\)

\(\frac{1}{5\cdot9}=\frac{1}{4}\cdot(\frac{1}{5}-\frac{1}{9})\)

\(\frac{1}{9\cdot13}=\frac{1}{4}\cdot(\frac{1}{9}-\frac{1}{13})\)

\(\frac{1}{(4n-3)\cdot(4n+1)}=\frac{1}{4}\cdot(\frac{1}{4n-3}-\frac{1}{4n+1})\)

\(S = \frac{1}{4}\cdot(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{4n-3}-\frac{1}{4n+1})=\)

\(=\frac{1}{4}\cdot(1-\frac{1}{4n+1})=\frac{1}{4}\cdot\frac{4n}{4n+1}=\frac{n}{4n+1}\)

\(1-\frac{1}{4n+1}=\frac{4n+1-1}{4n+1}=\frac{4n}{4n+1}\)

Ҷавоб: \(\frac{4n}{4n+1}\)