Q. The coefficient of $x^{n}$ in the polynomial $\left( x + \ ^{n}C_{0} \right)\left( x + {\ 3\ }^{n}C_{1} \right)\left( x + {\ 5\ }^{n}C_{2} \right)\ldots\lbrack x + (2n + 1)\ ^{n}C_{n}\rbrack$
A) $n.2^{n}$
B) $n.2^{n + 1}$
C) $(n + 1)2^{n}$
D) $n.2^{n} + 1$
Explanation
(c)
There are total $(n + 1)$ factors, let $P(x) = 0$
Let $\left( x + \ ^{n}C_{0} \right)\left( x + 3\ ^{n}C_{1} \right)\left( x + 5\ ^{n}C_{2} \right)\ldots\left\lbrack x + (2n + 1)\ ^{n}C_{n} \right\rbrack$
$= a_{n}x^{n} + a_{n - 1}x^{n - 1} + ... + a_{1}x + a_{0}$
Clearly, $a_{n} = 1$
and roots of the equation $P(x) = 0$ are $- \ ^{n}C_{0},\ - 3\ ^{n}C_{1},\ldots$
Sum of roots $= {- a}_{n - 1}/a_{n}$
$= - \ ^{n}C_{0} - 3\ ^{n}C_{1} - 5\ ^{n}C_{2}\ldots$
$\Rightarrow \ a_{n - 1} = (n + 1)2^{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.