A) \[\frac{{{3}^{n}}(2n+1)+1}{2({{3}^{n}})}\]
B) \[\frac{{{3}^{n}}(2n+1)-1}{2({{3}^{n}})}\]
C) \[\frac{{{3}^{n}}n-1}{2({{3}^{n}})}\]
D) \[\frac{{{3}^{n}}-1}{2}\]
Correct Answer: B
Solution :
The given series can be rewritten as \[\left( 1+\frac{1}{3} \right)+\left( 1+\frac{1}{9} \right)+\left( 1+\frac{1}{27} \right)+...n\] \[=n+\frac{1}{3}\left( 1+\frac{1}{3}+\frac{1}{{{3}^{2}}}+...n\,\,terms \right)\] \[=n+\frac{1}{3}\frac{\left[ 1-{{\left( \frac{1}{3} \right)}^{n}} \right]}{1-\frac{1}{3}}=n+\frac{\frac{1}{3}\left( 1-\frac{1}{{{3}^{n}}} \right)}{\frac{2}{3}}\] \[=\frac{2n+1-\frac{1}{{{3}^{n}}}}{2}=\frac{{{3}^{n}}(2n+1)-1}{2\cdot {{3}^{n}}}\]
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