If (1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + do | vedclass

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(A) Given $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$  --- $(1)$To find the sum of coefficients with even indices,we substitute $x

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$\frac{1}{2} (3^n + 1)$

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(A) Given $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$  --- $(1)$To find the sum of coefficients with even indices,we substitute $x = 1$ and $x = -1$ into equation $(1)$:For $x = 1$: $(1 - 1 + 1^2)^n = a_0 + a_1 + a_2 + \dots + a_{2n} \implies 1^n = a_0 + a_1 + a_2 + \dots + a_{2n} \implies 1 = a_0 + a_1 + a_2 + \dots + a_{2n}$ --- $(2)$For $x = -1$: $(1 - (-1) + (-1)^2)^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n} \implies (1 + 1 + 1)^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n} \implies 3^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n}$ --- $(3)$Adding equations $(2)$ and $(3)$:$1 + 3^n = (a_0 + a_1 + a_2 + \dots + a_{2n}) + (a_0 - a_1 + a_2 - a_3 + \dots + a_{2n})$$1 + 3^n = 2(a_0 + a_2 + a_4 + \dots + a_{2n})$Therefore,$a_0 + a_2 + a_4 + \dots + a_{2n} = \frac{3^n + 1}{2}$

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + \dots + a_{2n}$ is equal to

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