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JEE Main & Advanced Mathematics Sequence & Series Question Bank nth term of special series, Sum to n terms and Infinite number of terms

  • question_answer
    \[\frac{\frac{1}{2}.\frac{2}{2}}{{{1}^{3}}}+\frac{\frac{2}{2}.\frac{3}{2}}{{{1}^{3}}+{{2}^{3}}}+\frac{\frac{3}{2}.\frac{4}{2}}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}+.....n\] terms = [EAMCET 2000]

    A) \[{{\left( \frac{n}{n+1} \right)}^{2}}\]

    B) \[{{\left( \frac{n}{n+1} \right)}^{3}}\]

    C) \[\left( \frac{n}{n+1} \right)\]

    D) \[\left( \frac{1}{n+1} \right)\]

    Correct Answer: C

    Solution :

    \[{{T}_{n}}=\frac{\frac{n(n+1)}{2.\,2}}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.....+{{n}^{3}}}=\frac{\frac{n(n+1)}{4}}{{{\left( \frac{n(n+1)}{2} \right)}^{2}}}\] \[=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\] \ \[{{S}_{n}}=\sum\limits_{{}}^{{}}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}\] \[=\left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+.......+\left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[=1-\frac{1}{n+1}=\frac{n}{n+1}\].


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