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BCECE Engineering BCECE Engineering Solved Paper-2011

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

    A)  \[\frac{n-1}{2}\]                  

    B)                         \[\frac{n}{n+1}\]                 

    C)         \[\frac{n+1}{n+2}\]                            

    D)         \[\frac{(n+1)}{n}\]

    Correct Answer: B

    Solution :

    \[{{T}_{n}}=\frac{\frac{n(n+1)}{2.2}}{\sum {{n}^{3}}}=\frac{\frac{n(n+1)}{4}}{\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}}\] \[=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\]                 \[\therefore \]      \[{{S}_{n}}=\sum \left( \frac{1}{n}-\frac{1}{n+1} \right)\]                 \[=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\]                 \[=1-\frac{1}{n+1}=\frac{n}{n+1}\]


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