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

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