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sir this is the doubt : sigma r=0 to n (r+1)(nCr)^2. = ? some more are there. sigma r=1 to n (nCr)^2=?.

sir this is the doubt : sigma r=0 to n   (r+1)(nCr)^2. = ?  some more are there. 
sigma r=1 to n  (nCr)^2=?.

Grade:12

1 Answers

Yash Patari
askIITians Faculty 2123 Points
4 months ago
dear student,

∑r=0n​(r+1)Cr2​
It can be written as
∑r=0n​r⋅Cr2​+∑r=0n​Cr2​
We know that
r⋅nCr​=n⋅n−1Cr−1​
Hence
r⋅Cr2​=n⋅n−1Cr−12​
∑r=0n​n⋅n−1Cr−12​=n[n−1C02​+n−1C12​+.......+n−1Cn−12​]=n⋅(n−1)!(2n−1−n+1)!(2n−1)!​=n⋅(n−1)!n!(2n−1)!​−−−−(1)
∑r=0n​Cr2​=n!n!2n!​−−−−−(2)
Adding both eq(1) and (2)
∑r=0n​(r+1)Cr2​=n⋅(n−1)!n!(2n−1)!​+n!n!2n!​

Adding both eq(1) and (2)
∑r=0n​(r+1)Cr2​=n⋅(n−1)!n!(2n−1)!​+n!n!2n!​
∑r=0n​(r+1)Cr2​=n⋅(n−1)!n!(2n−1)!​+n(n−1)!n!2n(2n−1)!​
∑r=0n​(r+1)Cr2​=(n−1)!n!(n+2)(2n−1)!​

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