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Question

# If (1+x)n=n∑r=0 nCr xn, then C01⋅222+C12⋅323+C23⋅424+⋯+Cn(n+1)(n+2)2n+2 is equal to

A
3n+2+2n5(n+1)(n+2)
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B
3n+22n+5(n+1)(n+2)
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C
3n+22n5(n+1)(n+2)
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D
3n+2+2n+5(n+1)(n+2)
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Solution

## The correct option is C 3n+2−2n−5(n+1)(n+2) C01⋅222+C12⋅323+C23⋅424+⋯+Cn(n+1)(n+2)2n+2 =n∑r=01(r+1)(r+2) nCr⋅2r+2 =1(n+1)(n+2)n∑r=0n+2r+2⋅n+1r+1 nCr 2r+2 =1(n+1)(n+2)n∑r=0 n+2Cr+2 2r+2 Putting r+2=s =1(n+1)(n+2)[n+2∑s=2 n+2Cs2s] =1(n+1)(n+2)[(n+2∑s=0 n+2Cs2s)−( n+2C0 20+ n+2C1 21)] =1(n+1)(n+2)[(1+2)n+2−{1+2(n+2)}] =3n+2−2n−5(n+1)(n+2)

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