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公式C_(n+1)~m=C_n~m+C_n~(m-1)的一个应用利用组合数性质公式C_(n+1)~m=C_n~m+C=_n~(m-1)可以求形如{n(n+1)…(n+k-1)}的数列的前n项和S_n。 [例1] 求和 S=1·2·3+2·3·4+…+n(n+1)(n+2) 解:1/3!S=1·2·3/3!+2·3·4·/3!…+n(n+1)(n+2)/3! =C_3~3+C_4~3+…+C_(n+2)~3=(C_4~4+C_4~3)+C_5~3+…+C_(n+2)~3 =(C_5~4+C_5~3)+C_6~3+…+C_(n+2)~3=…=C_(n+2)~4+C_(n+2)~3 =C_(n+3)~4=n(n+1)(n+2)(n+3)/4!,
An application of the formula C_(n+1)~m=C_n~m+C_n~(m-1) uses the combination number property formula C_(n+1)~m=C_n~m+C=_n~(m-1) ) You can find the first n terms and S_n of a series with the form of {n(n+1)...(n+k-1)}. [Example 1] Summation S=1·2·3+2·3·4+...+n(n+1)(n+2) Solution: 1/3!S=1·2·3/3!+ 2·3·4·/3!...+n(n+1)(n+2)/3! =C_3~3+C_4~3+...+C_(n+2)~3=(C_4~4+ C_4~3)+C_5~3+...+C_(n+2)~3 =(C_5~4+C_5~3)+C_6~3+...+C_(n+2)~3=...=C_(n +2)~4+C_(n+2)~3 =C_(n+3)~4=n(n+1)(n+2)(n+3)/4!,