If (1+x-2x^2)^6 = 1+a_1x+a_2x^2+ldots+a_{12}x | vedclass

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(B) Given the expansion: $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$ Let $f(x) = (1+x-2x^2)^6 = 1+a_1x+a_2x^2+a_3x^3+\ldots+a_{12}x^{12}$. Putt

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$31$

Vedclass · Answer

(B) Given the expansion: $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$ Let $f(x) = (1+x-2x^2)^6 = 1+a_1x+a_2x^2+a_3x^3+\ldots+a_{12}x^{12}$. Putting $x=1$: $f(1) = (1+1-2)^6 = 0^6 = 0$. So, $0 = 1+a_1+a_2+a_3+a_4+\ldots+a_{12}$  $(i)$ Putting $x=-1$: $f(-1) = (1-1-2(-1)^2)^6 = (-2)^6 = 64$. So, $64 = 1-a_1+a_2-a_3+a_4-\ldots+a_{12}$  $(ii)$ Adding $(i)$ and $(ii)$: $0+64 = (1+a_1+a_2+a_3+a_4+\ldots+a_{12}) + (1-a_1+a_2-a_3+a_4-\ldots+a_{12})$ $64 = 2 + 2(a_2+a_4+a_6+\ldots+a_{12})$ $62 = 2(a_2+a_4+a_6+\ldots+a_{12})$ $a_2+a_4+a_6+\ldots+a_{12} = 31$.

If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

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