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Question

If (1x3)n=nr=0arxr(1x)3n2r, then find ar, where nN

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Solution

(1x3)n = (1x)n.(1+x+x2)n
(1x)n.((1x)2+3x)n = (1x)n(nr=0nCr.(1x)2n2r.(3x)r)
(1x)n.((1x)2+3x)n = (nr=0nCr.3r.(1x)3n2r.(x)r)
Hence, ar=nCr.3r

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