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Sum of Coefficients of All Terms

if a0 , a...

Question

if $a_{0}, a_{1}, a_{2}$ ....be the coefficients in the expansion of (1 + x + x $^{2})^{n}$ in ascending powers of x , then prove that
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . . + (- 1)^{n - 1} a_{n - 1}^{2}$
$= \frac{1}{2} a_{n} (1 - (- 1)^{n} a_{n})$

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Solution

Now $a_{2 n} = a_{0}, a_{2 n - 1} = a_{1}$

$∴ 2 [a_{0}^{2} + a_{1}^{2} + . . . . + (- 1)^{n - 1} a_{n - 1}^{2}] = (- 1)^{n} a_{n}^{2} = a_{n}$

or $a_{0}^{2} + a_{1}^{2} + . . . . + (- 1)^{n - 1} {a_{n - 1}}^{2}$

$= \frac{1}{2} a_{n} [1 - (- 1)^{n} a_{n}]$

This prove the second part

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Similar questions

a_{0}, a_{1}, a_{2}, a_{3} . . . .

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a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . + (- 1)^{n - 1} a_{n - 1}^{2} = \frac{1}{2} a_{n} {1 - (- 1)^{n} a_{n}}

a_{0}, a_{1}, a_{2}, . . . . . .

be the coefficients in the expansion of

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a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . . + a_{2 a}^{2} = a_{n}

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