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Question

If (1+x+x2+x3+.xp)n=a0+a1X+a2x2+..+anpxnp, then
a0+a1+a2+.+anp=

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

The correct option is D (p+1)n
Given,
(1+x2+x3+........xp)n=(a0+a1x+a2x2+.........+anpxnp)

Substituting x=1 in the above expression we get the sum of the binomial coefficients as
(1+x+x2...xp)n
=(1+(1+1+...1))n (substituting x=1)
=(1+p(1))n
=(1+p)n
Hence answer is D

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