Question

If $(1 + x)^n = C_o +. C_1x + C_2x^2 + ... + C_nx^n,$ then the sum of the series $C_o + 3C_1 + 5C_2 + ..... + (2n + 1) C_n$ will be

• A. $n.2^n$
• B. $(n+1)2^n$
• C. $n.2^n+1$
• D. $0$

Answer 1

Answer: (B)

$C_{0} + 3C_{1} + 5C_{2} + ..... + \left(2n + 1\right)C_{n}$
$= \left(C_{0} + C_{1} + ..... + C_{n}\right) + 2C_{1} + 4C_{2} + ....+2nC_{n}$
$= 2^{n} + 2 \left(C_{1}+ 2C_{2} + ..... + n.C_{n}\right)$
$ = 2^{n}+2\left(C_{1}+2C_{2}+..... + n.C_{n}\right)$
$= 2^{n}+2\left(n+2\cdot \frac{n\left(n-1\right)}{2\,!}+3 \frac{n\left(n-1\right)\left(n-2\right)}{3\,!}+ ... + n \cdot 1 \right)$
$= 2^{n}+2\left(n+n\left(n-1\right)+ \frac{n\left(n-1\right)\left(n-2\right)}{2\,!}+ ... +n\right)$
$= 2^{n}+2n \left(1+\left(n-1\right)+\frac{\left(n-1\right)\left(n-2\right)}{2\,!}+ ... +1\right)$
$= 2^{n}+2n . \left(^{n-1}c_{0 }+\,{}^{n-1}c_{1}+\,{}^{n-1}c_{2}....+\,{}^{n-1}c_{n-1}\right)$
$= 2^{n} + 2n. 2^{n-1} = 2^{n} + n.2^{n} = 2^{n} \left(n + 1\right)$.