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Binomial Theorem for Any Index

Prove that co...

Question

Prove that coefficient of $x^{4}$ in (1 + x - $2 x^{2})^{6}$ is -45 and if complete expansion of the expression is $1 + a_{1} x + a_{2} x^{2} + . . . . + a_{1} 2 x^{12}$
Prove that $a_{2} + a_{4} + a_{6} + . . . + a_{12} = 31$

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Solution

${(1 + x - 2 x^{2})}^{6} = (1 - x)^{6} (1 + 2 x)^{6}$
$= (1 - 6 x + 15 x^{2} - 20 x^{3} + 15 x^{4} . . . . . .) \times [1 + 6 (2 x) + 15 (2 x)^{2} + 20 (2 x)^{3} + 15 (2 x)^{4} + . . . .]$
$^{6} C_{1} = 6,^{6} C_{2} = 15,^{6} C_{3} = 20,^{6} C_{4} = 15$ etc.
Multiply and the term of $x^{4}$ will be
$[1 \cdot 15 \cdot 2^{4} - 6 \cdot 20 \cdot 2^{3} + 15 \cdot 15 \cdot 2^{2} - 20 \cdot 6 \cdot 2 + 15 \cdot 1] x^{4}$
Hence the coefficient of $x^{4}$ is
$240 - 960 + 900 - 240 + 15 = - 60 + 15 = - 45$
${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . . . . . . . + a_{12} x^{12}$
Putting, x = 1, we get
$0 = 1 + a_{1} + a_{2} + . . . . + a_{12}$
Putting, x = -1 we get
$64 = 1 - a_{1} + a_{2} - . . . . + a_{12}$
Adding (1) and (2) , we get
$64 = 2 (1 + a_{2} + a_{4} + . . . .)$ $∴ a_{2} + a_{4} + a_{6} + . . . . . . + a_{12} = 31$

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(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

a_{2} + a_{4} + a_{6} + . . . . + a_{12} =

(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}

(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}

(1 - x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

\frac{a_{2}}{2^{2}} + \frac{a_{4}}{2^{4}} + \frac{a_{6}}{2^{6}} + . . . . + \frac{a_{12}}{2^{12}}

{(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{12} x^{12},

then the value of

a_{2} + a_{4} + a_{6} + \dots + a_{12}

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