• # question_answer ${{n}^{th}}$ term of the series$\frac{{{1}^{3}}}{1}+\frac{{{1}^{3}}+{{2}^{3}}}{1+3}+\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+......$ will be [Pb. CET 2000] A) ${{n}^{2}}+2n+1$ B) $\frac{{{n}^{2}}+2n+1}{8}$ C) $\frac{{{n}^{2}}+2n+1}{4}$ D) $\frac{{{n}^{2}}-2n+1}{4}$

Obviously ${{T}_{n}}=\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.........+{{n}^{3}}}{1+3+5+.........\text{upto}\ n\ \text{terms}}$ = $\frac{\Sigma {{n}^{3}}}{\frac{n}{2}[2+(n-1)2]}=\frac{1}{4}\frac{{{n}^{2}}{{(n+1)}^{2}}}{{{n}^{2}}}=\frac{1}{4}({{n}^{2}}+2n+1)$.