Question

The sum of $1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+.......$ upto n terms is
[MP PET 1982]

• A. $\frac{25}{16}-\frac{4n+5}{16\times 5^{n-1}}.$
• B. $\frac{3}{4}-\frac{2n+5}{16\times 5^{n+1}}.$
• C. $\frac{3}{7}-\frac{3n+5}{16\times 5^{n-1}}.$
• D. $\frac{1}{2}-\frac{5n+1}{3\times 5^{n+2}}.$

Answer 1

The correct option is A $\frac{25}{16}-\frac{4n+5}{16\times 5^{n-1}}.$

Given series, let $S_n=1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+.......+\frac{n}{5^{n-1}}$

$\frac{1}{5}{S}_n=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+.......+\frac{n}{5^n}$
Subtracting, $(1-\frac{1}{5})S_n=1+\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+.....+\frac{n}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=\frac{1-\frac{1}{5^n}}{\frac{4}{5}}-\frac{n}{5^n}\Rightarrow S_n=\frac{25}{16}-\frac{4n+5}{16\times 5^{n-1}}.$