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Q.

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++2n.a2n=

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a

a0

b

n.3n

c

n.3n

d

3n

answer is C.

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Detailed Solution

(1+x+x2)n=a0+a1x+a2x2+a3x3++a2nx2n

Differentiating both sides with respect to x

n(1+x+x2)n1(1+2x)=a1+2a2x+3a3x2++2n.a2nx2n1

put x=1

n(3)n1(3)=a1+2a2+3a3++2n.a2n

n.3n=a1+2a2+3a3++2na2n

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