If (1+X+X2)n=∑2nr=0arxr then, a1−2a2+3a3....−2na2n is equal to
A
n
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B
−n
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C
0
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D
2n
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Solution
The correct option is B−n (1+x+x2)n=a0+a1x1+a2x2...+a2nx2n Differentiating with respect to x n(1+x+x2)n−1(2x+1)=a1+2a2x1...+2na2nx2n−1 Substituting x=−1 we get n(1−1+1)n−1(−2+1)=a1−2a2+3a3−4a4...+(−1)2n−12na2nx2n−1 =(n)(−1) =−n