A) \[\frac{7}{6}n+\frac{1}{6}-\frac{2}{{{3.2}^{n-1}}}\]
B) \[\frac{5}{3}n-\frac{7}{6}+\frac{2}{{{2.3}^{n-1}}}\]
C) \[n+\frac{1}{2}-\frac{1}{{{2.3}^{n}}}\]
D) \[n-\frac{1}{3}-\frac{1}{{{3.2}^{n-1}}}\]
Correct Answer: C
Solution :
Given series is\[1+\frac{4}{3}=+\frac{10}{9}+\frac{28}{27}+.....n\] terms \[=1+\left( 1+\frac{1}{3} \right)+\left( 1+\frac{1}{9} \right)+\left( 1+\frac{1}{27} \right)+.....n\] \[=(1+1+1+...+n\,\text{terms})\] \[+\left( \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....n\,\text{terms} \right)\] \[=n+\frac{\frac{1}{3}\left( 1-\frac{1}{{{3}^{n}}} \right)}{1-\frac{1}{3}}=n+\frac{1}{3}\times \frac{3}{2}[1-{{3}^{-n}}]\] \[=n+\frac{1}{2}[1-{{3}^{-n}}]=n+\frac{1}{2}-\frac{1}{{{2.3}^{n}}}\]
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