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) \[\frac{{{3}^{n}}-1}{2}\]
E) \[\frac{{{3}^{n}}-n}{2({{3}^{n}})}\]
Correct Answer: B
Solution :
\[\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.....\] up to n terms \[=\left( 1+\frac{1}{3} \right)+\left( 1+\frac{1}{9} \right)+\left( 1+\frac{1}{27} \right)+......\] upto n terms \[=n+\frac{1}{3}\left( 1+\frac{1}{3}+\frac{1}{{{3}^{2}}}+....+n\,terms \right)\] \[=n+\frac{\frac{1}{3}\left( 1-\frac{1}{{{3}^{n}}} \right)}{1-\frac{1}{3}}\] \[=\frac{2n+1-\frac{1}{{{3}^{n}}}}{2}\] \[=\frac{{{3}^{n}}(2n+1)-1}{2({{3}^{n}})}\]You need to login to perform this action.
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