If {(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{ | vedclass

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Question

(d) ${(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_{12}}{x^{12}}$.

Putting $x = 1$ and $x = -1$ and adding the results

$64 = 2(1+a_

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(d) ${(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_{12}}{x^{12}}$.

Putting $x = 1$ and $x = -1$ and adding the results

$64 = 2(1+a_2+a_4+...)$

$	herefore \,\,\,{a_2} + {a_4} + {a_6} + .... + {a_{12}} = 31$.

If ${(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_{12}}{x^{12}}$, then the expression ${a_2} + {a_4} + {a_6} + .... + {a_{12}}$ has the value

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