A) \[{{n}^{2}}+2n+1\]
B) \[\frac{{{n}^{2}}+2n+1}{8}\]
C) \[\frac{{{n}^{2}}+2n+1}{4}\]
D) \[\frac{{{n}^{2}}-2n+1}{4}\]
Correct Answer: C
Solution :
Obviously \[{{T}_{n}}=\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.........+{{n}^{3}}}{1+3+5+.........\text{upto}\ n\ \text{terms}}\] = \[\frac{\Sigma {{n}^{3}}}{\frac{n}{2}[2+(n-1)2]}=\frac{1}{4}\frac{{{n}^{2}}{{(n+1)}^{2}}}{{{n}^{2}}}=\frac{1}{4}({{n}^{2}}+2n+1)\].
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