Sequence And Series Question 114
Question: The sum of $ 1+\frac{2}{5}+\frac{3}{5^{2}}+\frac{4}{5^{3}}+……….. $ upto $ n $ terms is
[MP PET 1982]
Options:
A) $ \frac{25}{16}-\frac{4n+5}{16\times {5^{n-1}}} $
B) $ \frac{3}{4}-\frac{2n+5}{16\times {5^{n+1}}} $
C) $ \frac{3}{7}-\frac{3n+5}{16\times {5^{n-1}}} $
D) $ \frac{1}{2}-\frac{5n+1}{3\times {5^{n+2}}} $
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Answer:
Correct Answer: A
Solution:
Given series, let $ S_{n}=1+\frac{2}{5}+\frac{3}{5^{2}}+\frac{4}{5^{3}}+………+\frac{n}{{5^{n-1}}} $ $ \frac{1}{5}S_{n}=\text{ }\frac{1}{5}+\frac{2}{5^{2}}+\frac{3}{5^{3}}+…….+\frac{n}{5^{n}} $ Subtracting, $ ( 1-\frac{1}{5} )S_{n}=1+\frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}+……+upto\ n\ terms\ -\frac{n}{5^{n}} $
$ \Rightarrow $ $ \frac{4}{5}S_{n}=\frac{1-\frac{1}{5^{n}}}{\frac{4}{5}}-\frac{n}{5^{n}} $
$ \Rightarrow $ $ S_{n}=\frac{25}{16}-\frac{4n+5}{16\times {5^{n-1}}} $ .
BETA
