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Question

Consider (1+x+x2)2n=4nr=0arxr, where a0,a1,a2,...,a4n are real numbers and n is a positive integer. The value of a2 is

A
4n+1C2
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B
3n+1C2
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C
2n+1C2
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D
n+1C2
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Solution

The correct option is B 2n+1C2
We have (1+x+x2)2n =a0+a1x+a2x2...a4nx2n -----(i)
Hence a2 is the coefficient of x2
(1+(x+x2))2n
=1+2nC1(x+x2)+2nC2(x+x2)2 ------(ii)
Hence, coefficient of x2 is
2nC1+2nC2
=2n+(2n(2n1)2!)
=2n(2+2n12)
=((2n+1)(2n)2)
=2n+1C2

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