Question

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

• A. $\left(\frac{25}{16}\right)-\frac{\left(4n+5\right)}{\left(16·5^{n-1}\right)}$
• B. $\left(\frac{3}{4}\right)-\frac{\left(2n+5\right)}{\left(16·5^{n-1}\right)}$
• C. $\left(\frac{3}{7}\right)-\frac{\left(3n+5\right)}{\left(16·5^{n-1}\right)}$
• D. $\left(\frac{1}{6}\right)-\frac{\left(5n+1\right)}{\left(3·5^{n+2}\right)}$

Answer 1

The correct option is A

$\left(\frac{25}{16}\right)-\frac{\left(4n+5\right)}{\left(16·5^{n-1}\right)}$

Explanation for the correct option:

Let $S_n=1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{n}{5^{n-1}}...\left(1\right)$

Divide $S_n$ by $5$

$\frac{S_n}{5}=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{n}{5^n}...\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$

$\frac{4S_n}{5}=1+\frac{1}{5}+\frac{1}{5^2}+...-\frac{n}{5^n}$

$1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{n-1}}$ is a geometric progression having $a=1$and common difference $r=\frac{1}{5}$

Sum of geometric progression $S_n=a\left(\frac{r^n-1}{r-1}\right)$

Sum of geometric progression $1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{n-1}}$$=1\frac{\left[{\left({\displaystyle \frac{1}{5}}\right)}^n-1\right]}{{\displaystyle \frac{1}{5}-1}}$

$$\begin{gathered} =\frac{{\left({\displaystyle \frac{1}{5}}\right)}^n-1}{-{\displaystyle \frac{4}{5}}} \\ =\frac{5\left[1-{\left({\displaystyle \frac{1}{5}}\right)}^n\right]}{{\displaystyle 4}} \\ =\frac{1}{4}\left(\frac{5^n-1}{5^{n-1}}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{4S_n}{5}=1+\frac{1}{5}+\frac{1}{5^2}+..............-\frac{n}{5^n} \\ \Rightarrow \frac{4}{5}{S}_n=\frac{5\left[1-{\left({\displaystyle \frac{1}{5}}\right)}^n\right]}{4}-\frac{n}{5^n} \\ \Rightarrow S_n=\frac{5}{4}\left(\frac{5\left[1-{\left({\displaystyle \frac{1}{5}}\right)}^n\right]}{4}-\frac{n}{5^n}\right) \\ \Rightarrow S_n=\frac{25\left[1-{\left({\displaystyle \frac{1}{5}}\right)}^n\right]}{16}-\frac{5n}{4·5^n} \\ \Rightarrow S_n=\frac{25\left[5^n-1\right]}{16·5^n}-\frac{5n}{4·5^n} \\ \Rightarrow S_n=\frac{5}{4·5^n}\left(\frac{5\left[5^n-1\right]}{4}-n\right) \\ \Rightarrow S_n=\frac{1}{4·5^{n-1}}\left(\frac{5^{n+1}-5-4n}{4}\right) \\ \Rightarrow S_n=\left(\frac{5^{n+1}-(5+4n)}{16·5^{n-1}}\right) \\ \Rightarrow S_n=\left(\frac{5^{n+1}}{16·5^{n-1}}\right)-\frac{(4n+5)}{16·5^{n-1}} \\ \Rightarrow S_n=\frac{25}{16}-\frac{(4n+5)}{\left(16·5^{n-1}\right)} \end{gathered}$$

Hence, option (A) is correct.