Question

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

• A. $\frac{25}{16}-\frac{\left(4n+5\right)}{16\times 5^{n-1}}$
• 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. $\frac{1}{2}-\frac{5n+1}{3\times 5^{n-1}}$

Answer 1

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

$S=1+\frac{2}{5}+\frac{3}{5^2}+....+\frac{n}{5^{n-1}}\left(nterms\right)\frac{S}{5}=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{n-1}{5^{n-1}}+\frac{n}{5^n}\left(nterms\right)S-\frac{S}{5}=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+....+\frac{n-1}{5^{n-1}}-\frac{n}{5^n}\frac{4S}{5}=\frac{1(1-\frac{1}{5^n})}{1-\frac{1}{5}}-\frac{n}{5^n}\frac{4S}{5}=\frac{5^n-1}{4.5^{n-1}}-\frac{n}{5^n}\frac{4S}{5}=\frac{5}{4}-\frac{5}{4.5^n}-\frac{n}{5^n}S=\frac{25}{16}-\frac{(5+4n)}{16.5^{n-1}}$