Question

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

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

Hint:

Arithmetico-geometric series: use \(S_n - rS_n\) method.

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ T_n = \frac{3n-2}{5^{n-1}} \]
Step 2: Calculation / Simplification}
\[ S_n = \sum_{k=1}^n \frac{3k-2}{5^{k-1}} = 3\sum_{k=1}^n \frac{k}{5^{k-1}} - 2\sum_{k=1}^n \frac{1}{5^{k-1}} \]
\[ \sum_{k=1}^n \frac{1}{5^{k-1}} = \frac{1 - (1/5)^n}{1 - 1/5} = \frac{5}{4}\left(1 - \frac{1}{5^n}\right) \]
\[ \sum_{k=1}^n \frac{k}{5^{k-1}} = \frac{25}{16} - \frac{5n+4}{16 \cdot 5^{n-1}} \]
\[ S_n = 3\left[\frac{25}{16} - \frac{5n+4}{16 \cdot 5^{n-1}}\right] - 2\left[\frac{5}{4}\left(1 - \frac{1}{5^n}\right)\right] \]
\[ = \frac{5}{4} + \frac{15}{16}\left(1 - \frac{1}{5^{n-1}}\right) - \frac{3n-2}{4 \cdot 5^{n-1}} \]
Step 3: Final Answer
\[ \frac{5}{4} + \frac{15}{16}\left(1 - \frac{1}{5^{n-1}}\right) - \frac{3n-2}{4 \cdot 5^{n-1}} \]