证明对所有正整数 r, k 及 n (其中 n ≥ r) :
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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
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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