Q. If ${(1 + x)}^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ...C_{n}x^{n}$, then $C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + ... + C_{n}^{2}$ is equal to

A) $\frac{n!}{n!n!}$
B) $\frac{(2n)!}{n!n!}$
C) $\frac{(2n)!}{n!}$
D) None of these
Explanation

(b)
We have,
${(1 + x)}^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ... + C_{n}x^{n}$ …(i)
and $\left( 1 + \frac{1}{x} \right)^{n} = C_{0} + C_{1}\frac{1}{x} + C_{2}\left( \frac{1}{x} \right)^{2} + ... + C_{n}\left( \frac{1}{x} \right)^{n}$…(ii)
On multiplying Eqs. (i) and (ii) and taking the coefficient of constant terms in right hand side
$= C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + ... + C_{n}^{2}$
In right hand side ${(1 + x)}^{n}\left( 1 + \frac{1}{x} \right)^{n}$ or in $\frac{1}{x^{n}}{(1 + x)}^{2n}$ or term containing $x^{n}$ in ${(1 + x)}^{2n}$. Clearly the coefficient of $x^{n}$ in ${(1 + x)}^{2n}$ is equal to $\ ^{2n}C_{n} = \frac{(2n)\ !}{n\ !n\ !}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.