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if ${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r},$ then $a_{1} - 2 a_{2} + 3 a_{3} \dots - 2 {na}_{2 n}$ is equal to

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detailed solution

Correct option is B

We have, 1+x+x2n=a0+a1x+a2x2+a3x3+…+a2nx2nOn differentiating both sides, we get n1+x+x2n−1(1+2x)=a1+2a2x+3a3x2+...…+2na2nx2n−1 On putting x = -1, we get n(1−1+1)n−1(1−2)=a1−2a2+3a3−…−2na2n⇒a1−2a2+3a3−…−2na2n=−n

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if 1+x+x2n=∑r=02n arxr, then a1−2a2+3a3…−2na2n is equal to

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if ${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r},$ then $a_{1} - 2 a_{2} + 3 a_{3} \dots - 2 {na}_{2 n}$ is equal to