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Question

Coefficient of xk,(0kn) in expansion of P=1+(1+x)+(1+x)2......+(1+x)n

A
nCk
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B
n+1Cnk1
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C
nCnk
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D
n+1Ck+1
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Solution

The correct option is D n+1Ck+1
Given, P=1+(1+x)+(1+x)2..........+(1+x)n
On summing up the G.P., we get
P=(1+x)n+11x+11=(1+x)n+11x
Hence, to find out the coefficient xk we find coefficient of xk+1 as there is x in denominator.
So, coefficient is n+1Ck+1.

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