Evaluate ∑n,k1(nk) where the summation ranges over all positive integers n, k with 1<k<n−1

siutulysr5

siutulysr5

Answered

2022-01-23

Evaluate
n,k1(nk)
where the summation ranges over all positive integers n, k with 1<k<n1

Answer & Explanation

Frauental91

Frauental91

Expert

2022-01-24Added 15 answers

Let l=nk, we have
n=4k=2n2(nk)1=k=2n=k+2(nk)1
=k=2l=2(k+lk)1=k=2l=2k!l!(k+l)!
=k=2l=2(k+l+1)Γ(k+1)Γ(l+1)Γ(k+l+2)
=k=2l=2(k+l+1)01tk(1t)ldt
=01k=2l=2[(k+l+1)tk(1t)l]dt
Notice when s and t are independent, we have
k=2l=2(k+l+1)tksl=k=2l=2(tddt+sdds+1)tksl
=(tddt+sdds+1)t2s2(1t)(1s)
=s2t2(54(s+t)+3st)(1s)
Aaron Hughes

Aaron Hughes

Expert

2022-01-25Added 13 answers

Partial Fractions and Telescoping Sums
change order of summation
n=4k=2n21(nk)=k=2n=k+21(nk)
=k=2kk1n=k+2(1(n1k1)1(nk1))
Telescoping sum
=k=2kk11(k+1k1)
=k=22(k1)(k+1)
Partial fractions k=2(1k11k+1)
Telescoping sum
=1+12
Simplify
=32

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