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Question

2.nC0+22.nC12+23.nC23++2n+1.nCnn+1=

A
3n+112(n+1)
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B
3n+11n+1
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C
3n1n+1
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D
3n+1n+1
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Solution

The correct option is B 3n+11n+1
We have,
(1+x)n=nC0+nC1(x)+nC2(x)2+...+nCn1(x)n1+nCn(x)n
Integrating wrt. x,
(1+x)n+1n+1=nC0(x)+nC1(x)22+nC2(x)33+...+nCn(x)n+1n+1+k
Substitute x=0
1n+1=k
Substitute x=2,
3n+1n+1=nC0(2)+nC1(2)22+nC2(2)33+...+nCn(2)n+1n+1+1n+1
Hence,
nC0(2)+nC1(2)22+nC2(2)33+...+nCn(2)n+1n+1=3n+11n+1
Hence, option B is the correct answer.

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