If (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + ... | vedclass

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

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) Given the expansion: $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$.Let $f(x) = (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + a_3x^3 + .... + a_{12}x^{12}$.For $x = 1$: $f(1) = (1 + 1 - 2)^6 = 0^6 = 0 = 1 + a_1 + a_2 + a_3 + .... + a_{12}$.For $x = -1$: $f(-1) = (1 - 1 - 2(-1)^2)^6 = (-2)^6 = 64 = 1 - a_1 + a_2 - a_3 + .... + a_{12}$.Adding the two equations:$f(1) + f(-1) = 0 + 64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})$.$64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})$.$32 = 1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}$.Therefore,$a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 32 - 1 = 31$.

If $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$,then the expression $a_2 + a_4 + a_6 + .... + a_{12}$ has the value

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