If (1-x+x^2)^n = a_0 + a_1 x + ldots + a_{2n} | vedclass

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(D) Given the expansion: $(1-x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.Step $1$: Put $x = 1$ in the expansion:$(1 - 1 + 1)^n = a_0 +

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

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(D) Given the expansion: $(1-x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.Step $1$: Put $x = 1$ in the expansion:$(1 - 1 + 1)^n = a_0 + a_1 + a_2 + \ldots + a_{2n}$$1^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \Rightarrow 1 = a_0 + a_1 + a_2 + \ldots + a_{2n} \quad (i)$Step $2$: Put $x = -1$ in the expansion:$(1 - (-1) + (-1)^2)^n = a_0 + a_1(-1) + a_2(-1)^2 + \ldots + a_{2n}(-1)^{2n}$$(1 + 1 + 1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}$$3^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \quad (ii)$Step $3$: Add equations $(i)$ and $(ii)$:$1 + 3^n = (a_0 + a_1 + a_2 + \ldots + a_{2n}) + (a_0 - a_1 + a_2 - \ldots + a_{2n})$$1 + 3^n = 2(a_0 + a_2 + a_4 + \ldots + a_{2n})$Therefore, $a_0 + a_2 + a_4 + \ldots + a_{2n} = \frac{3^n + 1}{2}$.

If $(1-x+x^2)^n = a_0 + a_1 x + \ldots + a_{2n} x^{2n}$, then the value of $a_0 + a_2 + a_4 + \ldots + a_{2n}$ is

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