A) \[n(n-1)\]
B) \[n(n+1)\]
C) \[{{n}^{2}}\]
D) \[{{(n+1)}^{2}}\]
Correct Answer: C
Solution :
[c] \[{{S}_{n}}=1+2a+3{{a}^{2}}+......+n{{a}^{n-1}}\] or \[{{S}_{n}}=a+2{{a}^{2}}+......+(n-1){{a}^{n-1}}+n{{a}^{n}}\] where, \[a=1+\frac{1}{n}\] \[\therefore \,\,\,\,(1-a){{S}_{n}}=1+a+{{a}^{2}}+.....+{{a}^{n-1}}-n{{a}^{n}}\] \[(1-a){{S}_{n}}=\frac{{{a}^{n}}-1}{a-1}-n{{a}^{n}}\] \[\Rightarrow \,\,-\frac{1}{n}{{S}_{n}}=\frac{{{\left( 1+\frac{1}{n} \right)}^{n}}-1}{\frac{1}{n}}-n{{\left( 1+\frac{1}{n} \right)}^{n}}=-n\] \[\Rightarrow \,\,\,{{S}_{n}}={{n}^{2}}\]
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