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Question

If 2nr=0ar(x100)r=2nr=0br(x101)r and ak=2kkCn for all kn, then bn equals

A
2n(2n+11)
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B
2n(2n+1)
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C
2n(2n1)
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D
2n+1(2n1)
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Solution

The correct option is A 2n(2n+11)
Substitute x101=t, so that
2nr=0brtr=2nr=0ar(t+1)r ....(1)
bn= coefficient tn on the R.H.S of (1)
=nCnan+n+1Cnan+1+.....+2nCna2n
=2nk=nkCrak=2nk=n2k
=2n2nk=n2kn=2n(2n+11)

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