證明對所有正整數 r, k 及 n (其中 n ≥ r)，
C(n, r - 1) + C(n + 1, r - 1) + C(n + 2, r - 1) + ... + C(n + k, r - 1) = C(n + k + 1, r) - C(n, r)
THX !!!
因 nCr = (n+1)C(r+1) - nC(r+1) , 所以
nC(r-1) + (n+1)C(r-1) + (n+2)C(r-1) + ... + (n+k)C(r-1)
= [(n+1)Cr - nCr ] + [(n+2)Cr - (n+1)Cr] + [(n+3)Cr - (n+2)Cr] + ... + [(n+k)Cr - (n+k-1)Cr] + [ (n+k+1)Cr - (n+k)Cr]
= (n+k+1)Cr - nCr
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因 nCr = (n+1)C(r+1) - nC(r+1) , 所以
nC(r-1) + (n+1)C(r-1) + (n+2)C(r-1) + ... + (n+k)C(r-1)
= [(n+1)Cr - nCr ] + [(n+2)Cr - (n+1)Cr] + [(n+3)Cr - (n+2)Cr] + ... + [(n+k)Cr - (n+k-1)Cr] + [ (n+k+1)Cr - (n+k)Cr]
= (n+k+1)Cr - nCr