Binomial Theorem And Its Simple Applications Question 23
Question: If ’n’ is positive integer and three consecutive coefficient in the expansion of $ {{(1+x)}^{n}} $ are in the ratio 6 : 33 : 110, then n is equal to:
Options:
A) 9
B) 6
C) 12
D) 16
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Let the consecutive coefficient of
$ {{(1+x)}^{n}} $ are $ ^{n}{C_{r-1}},{{,}^{n}}C_{r},{{,}^{n}}{C_{r+1}} $
From the given condition, $ ^{n}{C_{r-1}}:{{,}^{n}}C_{r}:{{,}^{n}}{C_{r+1}}=6 $ $ :33:110 $
Now $ ^{n}{C_{r-1}}:{{,}^{n}}C_{r}=6:33 $
$ \Rightarrow \frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n-r)!}{n!}=\frac{6}{33} $
$ \Rightarrow \frac{r}{n-r+1}=\frac{2}{11}\Rightarrow 11r=2n-2r+2 $
$ \Rightarrow 2n-13r+2=0…(i) $
and $ ^{n}C_{r}:{{,}^{n}}{C_{r+1}}=33:110 $
$ \Rightarrow \frac{n!}{r!(n-r)!}\times \frac{(r+1)!(n-r-1)}{n!}=\frac{33}{110}=\frac{3}{10} $
$ \Rightarrow \frac{(r+1)}{n-r}=\frac{3}{10}\Rightarrow 3n-13r-10=0..(ii) $
Solving (i) & (ii), we get $ n=12 $
BETA
