Binomial Theorem And Its Simple Applications Question 204
Question: If $ \sum\limits_{r=0}^{n}{\frac{r+2}{r+1}{{,}^{n}}C_{r}=\frac{2^{8}-1}{6}} $ , then n is
Options:
A) 8
B) 4
C) 6
D) 5
Show Answer
Answer:
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 $
BETA
