If ${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2n}{a_{r}x^{r}},\ then\ a_{1} - 2a_{2} + 3a_{3}\ldots - 2na_{2n}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
We have, ${(1 + x + x^{2})}^{n} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ... + a_{2n}x^{2n}$
On differentiating both sides, we get
$n\left( 1 + x + x^{2} \right)^{n - 1}(1 + 2x) = a_{1} + 2a_{2}x + 3a_{3}x^{2} + ... + 2na_{2n}x^{2n - 1}$
On putting $x = - 1$, we get
$n(1 - 1 + 1)^{n - 1}(1 - 2) = a_{1} - 2a_{2} + 3a_{3} - ... - 2na_{2n}$
$\Rightarrow \ a_{1} - 2a_{2} + 3a_{3} - ... - 2na_{2n} = - n$

Q. If ${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2n}{a_{r}x^{r}},\ then\ a_{1} - 2a_{2} + 3a_{3}\ldots - 2na_{2n}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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