If the sum of the coefficients of even powers | vedclass

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(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 $n=$

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