Question

The ratio of coefficients of 9^th and 7^th terms in the expansion of $( 1 + x)^n$ is 9 : 7 . Then the coefficient of 4^th term is

• A. 395
• B. 455
• C. 530
• D. 645

Answer 1

Answer: (B)

We have, $(1 + x)^n$
$T_{9} = \,{}^{n}C_{8} x^{8} $
$ T_{7} = \,{}^{n}C_{6}x^{6} $
Given, $\,{}^{n}C_{8} :\,{}^{n}C_{6} = 9 : 7 $
$\Rightarrow \frac{^{n}C_{8}}{^{n}C_{6}} = \frac{9}{7}$
$\Rightarrow \frac{n! 6!\left(n-6\right)!}{8!\left(n-8\right)! n!} = \frac{9}{7}$
$ \frac{\left(n-6\right)\left(n-7\right)}{8\times 7} = \frac{9}{7}$
$ \Rightarrow \left(n-6\right)\left(n-7\right) = 9\times 8 $
$ n =15$
Coefficient of $4$th term of $\left(1+x\right)^{15}$
$\,{}^{15}C_{3} = \frac{15!}{3!12!} = 455 $