If $(1 + x + x^{2})^{n} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ , then $2 a_{1} - 3 a_{2} + . . . - (2 n + 1) a_{2 n}$ is equal to

n

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- n

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n + 1

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- n - 1

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- n + 1

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Solution

The correct option is C $- n$
Given,
$(1 + x + x^{2})^{n} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$

$\Rightarrow x (1 + x + x^{2})^{n} = x + a_{1} x^{2} + a_{2} x^{3} + . . . + a_{2 n} x^{2 n + 1}$
On diffferentiating w.r.t. $x$ , we get
$(1 + x + x^{2})^{n} + x \cdot n (1 + x + x^{2})^{n - 1} (1 + 2 x)$ $= 1 + 2 a_{1} x + 3 a_{2} x^{2} + . . . + a_{2 n} \cdot (2 n + 1) x^{2 n}$

On putting $x = - 1$ , we get

$(1 - 1 + 1)^{n} - n (1 - 1 + 1)^{n - 1} (1 - 2)$ $= 1 - 2 a_{1} + 3 a_{2} + . . . + a_{2 n} (2 n + 1)$

$\Rightarrow 1 - n (- 1) = 1 - 2 a_{1} + 3 a_{2} + . . . + a_{2 n} (2 n + 1)$

$\Rightarrow 2 a_{1} - 3 a_{2} . . . . = (2 n + 1) a_{2 n} = - n$

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If (1+x+x2)n=1+a1x+a2x2+...+a2nx2n, then 2a1−3a2+...−(2n+1)a2n is equal to

If $(1 + x + x^{2})^{n} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ , then $2 a_{1} - 3 a_{2} + . . . - (2 n + 1) a_{2 n}$ is equal to