Q80 / 125 · BITSAT 2016 · Mathematics

If $\sum_{r=0}^{n} \frac{r+2}{r+1}^{n} C_{r}=\frac{2^{8}-1}{6},$ then $n=$

  • A
    8
  • B
    4
  • C
    6
  • Correct answer
    D
    5

Solution

$\sum_{r=0}^{n} \frac{r+2}{r+1}{ }^{n} C_{r}=\frac{2^{8}-1}{6}$ $$ \begin{aligned} & \Rightarrow \sum_{r=0}^{n}\left[1+\frac{1}{r+1}\right]{ }^{n} C_{r}=\frac{2^{8}-1}{6} \\ & \Rightarrow 2^{n}+\sum_{r=0}^{n} \frac{1}{n+1} \cdot{ }^{n+1} C_{r+1}=\frac{2^{8}-1}{6} \\ \Rightarrow 2^{n}+& \frac{2^{n+1}-1}{n+1}=\frac{2^{8}-1}{6} \Rightarrow \frac{2^{n}(n+3)-1}{n+1}=\frac{2^{8}-1}{6} \\ \Rightarrow & \frac{2^{n}(n+1+2)-1}{n+1}=\frac{2^{5}(6+2)-1}{6} \end{aligned} $$ Comparing we get $\mathrm{n}+1=6 \Rightarrow \mathrm{n}=5$

Hence, the correct answer is option: 'D'.

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