Q. If ${(1 + x)}^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ... + C_{n}x^{n}$, then the value of $C_{0} + 2C_{1} + 3C_{2} + ... + (n + 1)C_{n}$ will be
A) $(n + 2)2^{n - 1}$
B) $(n + 1)2^{n}$
C) $(n + 1)2^{n - 1}$
D) $(n + 2)2^{n}$
Explanation
(a)
Since, ${x(1 + x)}^{n} = xC_{0} + C_{1}x^{2} + C_{2}x^{3} + ... + C_{n}x^{n + 1}$
On differentiating w.r.t. $x$, we get
${(1 + x)}^{n} + nx{(1 + x)}^{n - 1} = C_{0} + 2C_{1}x + 3C_{2}x^{2} + ... + (n + 1)C_{n}x^{n}$
Put $x = 1,$ we get
$C_{0} + 2C_{1} + 3C_{2} + ... + (n + 1)C_{n} = 2^{n} + n2^{n - 1}$
$= 2^{n - 1}(n + 2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.