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Question

If 1+(1+x)+(1+x)2+(1+x)3+......+(1+x)n=nk=0akxk, then which of the following is true?


A

an2=n(n1)2

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B

a29a28= n+1C10(n+1C10n+1C9)

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C

ak= nCk

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D

nk=0ak=2n+11

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Solution

The correct option is D

nk=0ak=2n+11


1+(1+x)+(1+x)2+(1+x)3+......+(1+x)n=(1+x)n+11x=nk=0akxk

So, the coefficient of xk in (1+x)n+11x=ak=n+1Ck+1

an2=n+1Cn1=n(n+1)2

a29a28=n+2C10(n+1C10n+1C9)

nk=0ak=n+1C1+n+1C2+....+n+1Cn+1=2n+11


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