If the sum of the coefficients of even powers | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $(1-x+x^2)^{2n} = a_0 + a_1 x + a_2 x^2 + \dots + a_{4n} x^{4n}$.Put $x = 1$,we get $1^{2n} = 1 = a_0 + a_1 + a_2 + \dots + a_{4n} \dots (i)$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let $(1-x+x^2)^{2n} = a_0 + a_1 x + a_2 x^2 + \dots + a_{4n} x^{4n}$.Put $x = 1$,we get $1^{2n} = 1 = a_0 + a_1 + a_2 + \dots + a_{4n} \dots (i)$.Put $x = -1$,we get $(1 - (-1) + (-1)^2)^{2n} = 3^{2n} = a_0 - a_1 + a_2 - \dots + a_{4n} \dots (ii)$.Adding equations $(i)$ and $(ii)$,we get $1 + 3^{2n} = 2(a_0 + a_2 + a_4 + \dots + a_{4n})$.The sum of coefficients of even powers of $x$ is $a_0 + a_2 + \dots + a_{4n} = 3281$.Therefore,$1 + 3^{2n} = 2 	imes 3281 = 6562$.$3^{2n} = 6561$.Since $3^8 = 6561$,we have $2n = 8$,which implies $n = 4$.

If the sum of the coefficients of even powers of $x$ in the expansion of $(1-x+x^2)^{2n}$ is $3281$,then $n=$

Similar Questions

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

If $C_r = ^{100}C_r$,then the value of $1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$ is equal to:

If $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_3 + a_6 + \dots =$

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module