A) \[n(n-1)\]
B) \[n(n+1)\]
C) \[{{n}^{2}}\]
D) \[{{(n+1)}^{2}}\]
Correct Answer: C
Solution :
\[\sum\limits_{k\,=1}^{n}{k\,{{\left( 1+\frac{1}{n} \right)}^{k-1}}}\] \[=1+2{{\left( 1+\frac{1}{n} \right)}^{1}}+3{{\left( 1+\frac{1}{n} \right)}^{2}}+....\]upto n terms = \[1+2t+3{{t}^{2}}+...\]upto n terms = \[{{(1-t)}^{-2}}={{\left[ 1-\left( 1+\frac{1}{n} \right) \right]}^{-2}}\] = \[{{\left( \frac{1}{n} \right)}^{-2}}={{(n)}^{2}}={{n}^{2}}\].You need to login to perform this action.
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