Суммаро ҳисоб кунед:

\(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...\)

\(S=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...\)

\(5S=1+\frac{2}{5}+...+\frac{n}{5^n-1}+...\)

\(5S-S=1+\frac{2}{5}+...+\frac{n}{5^n-1}+...-(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...)\)

\(4S=1+(\frac{2}{5}-\frac{1}{5})+(\frac{3}{5^2}-\frac{2}{5^2})+...+\)

\(+(\frac{n}{5^{n-1}}-\frac{n-1}{5^{n-1}})+...=\)

\(=1+\frac{1}{5}+\frac{1}{5^2}+...++\frac{1}{5^{n-1}}+...\)

\(b_1=1;b_2=\frac{1}{5^2};b_3=\frac{1}{5^3};...\)

\(q=\frac{1}{5}\)

\(C=b_1+b_2+...+b_n+...=\frac{b_1}{1-q}=\)

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

\(4S=C=\frac{5}{4}\)

\(S=\frac{5}{16}\)

Ҷавоб:\(\frac{5}{16}\)