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Question

2.C0+22.C12+23.C23+...+2n+1.Cnn+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 C 3n+1n+1
We know that (1+x)n=nC0+nC1x+nC2x2+....+....nCnxn
Integrating both sides we get;
(1+x)n+1n+1=nC0x+nC1x22+nC2x33+nC3x44+.....nCnxn+1n+1
Putting x=2, we get
(1+2)n+1n+1=2.nC0+22nC12+23nC23+........2n+1nCnn+1
3n+1n+1=2C0+22C12+23C23+....2n+1Cnn+1

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