If${(1 + x - {2x}^{2})}^{6} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{12}x^{12}$ , then the expression $a_{2} + a_{4} + a_{6} + ... + a_{12}$ has the value

Question

Dr. Abubacker · Accepted Answer

(d)
Given that, ${(1 + x - {2x}^{2})}^{6} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{12}x^{12}$
On putting $x = 1\ and\ x = - 1$ and adding the results, we get
$64 = 2(1 + a_{2} + a_{4} + ... + a_{12})$
$	herefore\ a_{2} + a_{4} + a_{6} + ... + a_{12} = 31$

Q. If${(1 + x - {2x}^{2})}^{6} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{12}x^{12}$ , then the expression $a_{2} + a_{4} + a_{6} + ... + a_{12}$ has the value

Explanation

Reviewed by Dr. Abubacker

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