Question

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

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

Answer 1

Answer: (A)

Clearly, the given series is an arithmetic geometric series whose corresponding $A P$ and $G P$ are respectively,
$1,4,7,10, \ldots$ and $1, \frac{1}{5}, \frac{1}{5^2}, \frac{1}{5^{3}}, \ldots$
The $n^{t h}$ term of $A P=[1+(n-1) \times 3]=3 n-2$
The $n^{t h}$ term of $G P=\left[1 \times\left(\frac{1}{5}\right)^{n-1}\right]=\left(\frac{1}{5}\right)^{n-1}$
So, the $n^{t h}$ term of the given series is
$(3 n-2) \times \frac{1}{5^{n-1}}=\frac{3 n-2}{5^{n-1}}$
Let, $S_{n}=1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\ldots+\frac{3 n-5}{5^{n-2}}+\frac{3 n-2}{5^{n-1}} \,\,\,...(i)$
$\frac{1}{5} S_{n}=\frac{1}{5}+\frac{4}{5^{2}}+\frac{7}{5^{3}}+\ldots+\frac{(3 n-5)}{5^{n-1}}+\frac{3 n-2}{5^{n}} \,\,\,...(ii)$
Subtracting (ii) from (i), we get
$S_{n}-\frac{1}{5} S_{n}=1+\left\{\frac{3}{5}+\frac{3}{5^{2}}+\frac{3}{5^{3}}+\ldots+\frac{3}{5^{n-1}}\right\}-\frac{(3 n-2)}{5^{n}}$
$\Rightarrow \frac{4}{5} S_{n}=1+\frac{3}{5} \frac{\left\{1-\left(\frac{1}{5}\right)^{n-1}\right\}}{\left(1-\frac{1}{5}\right)}-\frac{(3 n-2)}{5^{n}}$
$\Rightarrow \frac{4}{5} S_{n}=1+\frac{3}{5} \frac{\left\{1-\frac{1}{5^{n-1}}\right\}}{\left(\frac{4}{5}\right)}-\frac{(3 n-2)}{5^{n}}$
$\Rightarrow \frac{4}{5} S_{n}=1+\frac{3}{4}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3 n-2)}{5^{n}}$
$\Rightarrow S_{n}=\frac{5}{4}+\frac{15}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3 n-2)}{4\cdot 5^{n-1}}$