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JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Self Evaluation Test - Binomial Theorem

  • question_answer
    If \[\sum\limits_{r=0}^{n}{\frac{r+2}{r+1}{{\,}^{n}}{{C}_{r}}=\frac{{{2}^{8}}-1}{6}}\], then n is

    A) 8

    B) 4

    C) 6

    D) 5

    Correct Answer: D

    Solution :

    [d] \[\sum\limits_{r=0}^{n}{\frac{r+2}{r+1}{{\,}^{n}}{{C}_{r}}=\sum\limits_{r=0}^{n}{\frac{r+1+1}{r+1}{{\,}^{n}}{{C}_{r}}}}\]
    \[=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}+\sum\limits_{r=0}^{n}{\frac{{{\,}^{n}}{{C}_{r}}}{r+1}={{2}^{n}}+\sum\limits_{r=0}^{n}{\frac{^{n+1}{{C}_{r+1}}}{n+1}}}}\]
    \[={{2}^{n}}+\frac{1}{n+1}\sum\limits_{r+0}^{n}{^{n+1}{{C}_{r+1}}}\]
    \[={{2}^{n}}+\frac{1}{n+1}({{2}^{n+1}}-1)\]
    \[=\frac{1}{n+1}[(n+1){{2}^{n}}+{{2}^{n+1}}-1]\]
    \[=\frac{1}{n+1}[{{2}^{n}}(n+3)-1]\]
    Given, \[\frac{(n+3){{2}^{n}}-1}{n+1}=\frac{{{2}^{8}}-1}{6}=\frac{(5+3){{.2}^{5}}-1}{5+1}\]
    \[\Rightarrow \,\,\,\,n=5\]


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