Question

The sum upto $n$ terms of the series $1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\dots$ is

• A. $\frac{5}{4}+\frac{15}{16}(1-\frac{1}{5^{n-1}})-\frac{3n-2}{4\left(5^{n-1}\right)}$
• B. $\frac{4}{5}+\frac{16}{15}(1-\frac{1}{5^{n-1}})+\frac{3n-2}{5^{n-1}}$
• C. $\frac{4}{5}+\frac{16}{15}\frac{1}{5^{n-1}}+(1-\frac{3n-2}{5^{n-1}})$
• D. $\frac{5}{4}+\frac{15}{16}\frac{1}{5^{n-1}}-\frac{3n}{4\left(5^{n-1}\right)}$

Answer 1

The correct option is A $\frac{5}{4}+\frac{15}{16}(1-\frac{1}{5^{n-1}})-\frac{3n-2}{4\left(5^{n-1}\right)}$

$1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\dots$
Clearly, the given series is an AGP.
Let,
$S_n=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\dots +\frac{3n-5}{5^{n-2}}+\frac{3n-2}{5^{n-1}}...\left(1\right)$
$\frac{1}{5}{S}_n=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\dots +\frac{(3n-5)}{5^{n-1}}+\frac{3n-2}{5^n}...\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right),$ we get
$S_n-\frac{1}{5}{S}_n=1+[\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^2}+\dots +\frac{3}{5^{n-1}}]-\frac{(3n-2)}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=1+\frac{3}{5}\frac{(1-{\left(\frac{1}{5}\right)}^{n-1})}{(1-\frac{1}{5})}-\frac{(3n-2)}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=1+\frac{3}{5}\frac{[1-\frac{1}{5^{n-1}}]}{\left(\frac{4}{5}\right)}-\frac{(3n-2)}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=1+\frac{3}{4}(1-\frac{1}{5^{n-1}})-\frac{(3n-2)}{5^n}$
$\Rightarrow S_n=\frac{5}{4}+\frac{15}{16}(1-\frac{1}{5^{n-1}})-\frac{3n-2}{4\left(5^{n-1}\right)}$


Alternate solution:
For $n=1$,
$S=1$
Now, checking the options for $n=1$
$\frac{5}{4}+\frac{15}{16}(1-\frac{1}{5^{1-1}})-\frac{3-2}{4\left(5^{1-1}\right)}=1$

$\frac{4}{5}+\frac{16}{15}(1-\frac{1}{5^{1-1}})+\frac{3-2}{5^{1-1}}=\frac{9}{5}$

$\frac{4}{5}+\frac{16}{15}\frac{1}{5^{1-1}}+(1-\frac{3-2}{5^{1-1}})=4$

$\frac{5}{4}+\frac{15}{16}\frac{1}{5^{1-1}}-\frac{3}{4\left(5^{1-1}\right)}=\frac{23}{16}$