11th Class Mathematics Other Series Question Bank Critical Thinking

  • 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) \[\frac{{{n}^{2}}-2n+1}{4}\]

    Correct Answer: C

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner