Question

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

• A. $\frac{5}{4}+\frac{15}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n-2)}{4\cdot 5^{n-1}}$
• B. $\frac{5}{4}+\frac{1}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{3n}{4\cdot 5^{n-1}}$
• C. $\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n+2)}{4\cdot 5^{n-1}}$
• D. None of the above

Answer 1

Answer: (A)

Clearly, the given series is an arithmetic geometric series whose corresponding $AP$ and $GP$ are respectively,
$1,4,7,10,\dots$ and $1,\frac{1}{5},\frac{1}{5^2},\frac{1}{5^3},\dots$
The $n^{th}$ term of $AP=[1+(n-1)\times 3]=3n-2$
The $n^{th}$ term of $GP=\left[1\times {\left(\frac{1}{5}\right)}^{n-1}\right]={\left(\frac{1}{5}\right)}^{n-1}$
So, the $n^{th}$ term of the given series is
$(3n-2)\times \frac{1}{5^{n-1}}=\frac{3n-2}{5^{n-1}}$
Let, $S_n=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\dots +\frac{3n-5}{5^{n-2}}+\frac{3n-2}{5^{n-1}}...(i)$
$\frac{1}{5}{S}_n=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\dots +\frac{(3n-5)}{5^{n-1}}+\frac{3n-2}{5^n}...(ii)$
Subtracting (ii) from (i), we get
$S_n-\frac{1}{5}{S}_n=1+\left\{\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+\dots +\frac{3}{5^{n-1}}\right\}-\frac{(3n-2)}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=1+\frac{3}{5}\frac{\left\{1-{\left(\frac{1}{5}\right)}^{n-1}\right\}}{\left(1-\frac{1}{5}\right)}-\frac{(3n-2)}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=1+\frac{3}{5}\frac{\left\{1-\frac{1}{5^{n-1}}\right\}}{\left(\frac{4}{5}\right)}-\frac{(3n-2)}{5^n}$
$\Rightarrow \frac{4}{5}{S}_n=1+\frac{3}{4}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n-2)}{5^n}$
$\Rightarrow S_n=\frac{5}{4}+\frac{15}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n-2)}{4\cdot 5^{n-1}}$