$\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{40} x^{40}$. Then, $a_{1}+a_{3}+a_{5}+\ldots .+a_{37}$ is equal to :

(C) $2^{19}\left(2^{20}-21\right)$

(A) $2^{20}\left(20^{20}+21\right)$

(B) $2^{19}\left(2^{20}+21\right)$

(D) $2^{20}\left(2^{20}-21\right)$

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Binomial Theorem

(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . + a_{40} x^{40}

. Then,

a_{1} + a_{3} + a_{5} + \dots . + a_{37}

is equal to :

Answer: (C)

Let

P (x) = (1 + x + 2 x^{2})^{20}

. Then

P (1) = 4^{20} = 2^{40}

and

P (- 1) = (1 - 1 + 2)^{20} = 2^{20}

, so the sum of all odd-indexed coefficients is

a_{1} + a_{3} + \dots + a_{39} = \frac{P ( 1 ) - P ( - 1 )}{2} = \frac{2 ^{40} - 2 ^{20}}{2} = 2^{19} (2^{20} - 1)

The required sum stops at

a_{37}

, so

a_{39}

has to be removed. A term in

x^{39}

arises only by taking

x

from one factor and

2 x^{2}

from the remaining nineteen, giving

a_{39} = (1 20) 2^{19} = 20 \cdot 2^{19}

Subtracting,

a_{1} + a_{3} + \dots + a_{37} = 2^{19} (2^{20} - 1) - 20 \cdot 2^{19} = 2^{19} (2^{20} - 21)

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option