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Question

If (1+x)n=nr=0nCrxr and nr=01nCr=a, then the value of 0in 0jn(inCi+jnCj) is equal to

A
n2a
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B
n22a
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C
na2
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D
na(n+1)
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Solution

The correct option is D na(n+1)
Given, a=nr=01nCr=nr=01nCnr
Let,
y=0in 0jn(inCi+jnCj)
=0in 0jn(ninCni+njnCnj) [nr=0rnCr=nr=0nrnCnr]
=n0in 0jn(1nCni+1nCnj)y=n0in(n+1nCni+a)y
2y=n{a(n+1)+a(n+1)}
y=na(n+1)

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