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Question

If ak is the coefficient of xk in the expansion of (1+x+x2)n for k=0,1,2, , 2n, then a1+2a2+3a3++2na2n=

A
a0
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B
3n
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C
n.3n
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D
n3n
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Solution

The correct option is C n.3n
(1+x+x2)n=1+a1x+a2x2+a3x3...+a2nx2n
Differentiating the above expression with respect to x, we get
n(1+x+x2)n1(2x+1)=a1+2a2x+3a3x2+4a4x3...+2na2nx2n1
Let x=1,
Therefore, we get
n3n1(3)=a1+2a2+3a3...+2na2n
a1+2a2+3a3..+2na2n=n.3n

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