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

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(A) Given the expansion: $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$.Putting $x = 1$,we get:$(1 - 1 + 1)^n = a_0 + a_1 + a_2 + ....

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

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(A) Given the expansion: $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$.Putting $x = 1$,we get:$(1 - 1 + 1)^n = a_0 + a_1 + a_2 + .... + a_{2n}$$1 = a_0 + a_1 + a_2 + .... + a_{2n}$ ..... $(i)$Putting $x = -1$,we get:$(1 - (-1) + (-1)^2)^n = a_0 + a_1(-1) + a_2(-1)^2 + .... + a_{2n}(-1)^{2n}$$(1 + 1 + 1)^n = a_0 - a_1 + a_2 - .... + a_{2n}$$3^n = a_0 - a_1 + a_2 - .... + a_{2n}$ ..... $(ii)$Adding $(i)$ and $(ii)$,we get:$1 + 3^n = (a_0 + a_1 + a_2 + .... + a_{2n}) + (a_0 - a_1 + a_2 - .... + a_{2n})$$3^n + 1 = 2(a_0 + a_2 + a_4 + .... + a_{2n})$Therefore,$a_0 + a_2 + a_4 + .... + a_{2n} = \frac{3^n + 1}{2}$.

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + .... + a_{2n} = $

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