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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Binomial Theorem

BITSAT 2016 · Mathematics · Binomial Theorem

If ∑_ r=0 ^ n (r+2)/(r+1)^ n C_ r = 2^ 8 -1 6 , then n=

If ∑r=0nr+2r+1nCr=28−16,\sum_{r=0}^{n} \frac{r+2}{r+1}^{n} C_{r}=\frac{2^{8}-1}{6}, then n=n=

Options

  1. A8
  2. B4
  3. C6
  4. D5

Correct answer

D. 5

Step-by-step solution

∑r=0nr+2r+1(nr)=28−16\sum_{r=0}^{n} \frac{r+2}{r+1}\binom{n}{r}=\frac{2^{8}-1}{6}⇒∑r=0n[1+1r+1](nr)=28−16⇒2n+∑r=0n1n+1⋅(n+1r+1)=28−16⇒2n+2n+1−1n+1=28−16⇒2n(n+3)−1n+1=28−16⇒2n(n+1+2)−1n+1=25(6+2)−16\begin{aligned}& \Rightarrow \sum_{r=0}^{n}\left[1+\frac{1}{r+1}\right]\binom{n}{r}=\frac{2^{8}-1}{6}\\ & \Rightarrow 2^{n}+\sum_{r=0}^{n} \frac{1}{n+1} \cdot\binom{n+1}{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}$ComparingwegetComparing we get\mathrm{n}+1=6 \Rightarrow \mathrm{n}=5$
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