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设S_t(n)=1~t+2~t+…+n~t,SS_t(n)=sum from i=1 to n sum from j=1 to i (j~t)=1~t+(1~t+2~t)+…+(1~t+2~t+…+n~t),t∈N,并称SS_t(n)为自然数方幂的累进和。文[1]给出:
Let S_t(n)=1~t+2~t+...+n~t,SS_t(n)=sum from i=1 to n sum from j=1 to i (j~t)=1~t+(1~ t+2~t)+...+(1~t+2~t+...+n~t), t∈N, and said SS_t(n) is the progressive sum of powers of natural numbers. Text [1] gives: