1*2*3+2*3*4+……+n（n+1）（n+2）=?

1*2*3+2*3*4+……+n（n+1）（n+2）=?

n(n+1)(n+2)=n^3+3n^2+2n

1^3+2^3+......+n^3 =[n(n+1)/2]^2

1^2+2^2+.....+n^2=n(n+1)(2n+1)/6

1+2+3+...+n=n(n+1)/2

1*2*3+2*3*4+n(n+1)(n+2)=[n(n+1)/2]^2 +n(n+1)(2n+1)/2+n(n+1)

=n(n+1)/2(n(n+1)/2+2n+3)=n(n+1)(n^2+5n+6)/4=n(n+1)(n+2)(n+3)/4

什么玩意，先乘后加？？？

有： ① 2*(1+2+3+4+...+N)=N(N+1) ; ② 3*(1^2+2^2+3^2+...+N^2)=3*[N(N+1)(2N+1)/6] ; ③ 1^3+2^3+3^3+...+N^3=[N(N+1)/2]^2 ; 由①+②+③得： 1*2*3+2*3*4+······+N（N+1)(N+2) =N(N+1)(N+2)(N+3)/4

因为(n-1)*n*(n+1)=n^3-n 所以 1*2*3+2*3*4…+99*100*101 =2^3+3^3+4^3+…+100^3-2-3-4-…- 100 =1^3+2^3+3^3+4^3+…+100^3-1-2-3-4-…- 100 =(100*101/2)^2-5050 =25497450 （ 因为^3+2^3+3^3+4^3+…+n^3=(n*(n+1)/2)^2)

n(n+1)(n+2) =n(n²+3n+2) =n³+3n²+n ∴1×2×3+2×3×4+…n(n+1)(n+2) =(1³+3*1²+1)+(2³+3*2²+2)+……+(n³+3n²+n) =(1³+2³+……+n³)+3(1²+2²+……n²)+(1+2+……n) =[n(n+1)/2]^2+3[n(n+1)(2n+1)/6]+n(n+1)/2 =n²(n+1)²/4+n(n+1)(2n+1)/2+n(n+1)/2 =n(n+1)[n(n+1)+2(2n+1)+2]/4 =n(n+1)(n²+n+4n+2+2)/4 =n(n+1)(n²+5n+4)/4 =n(n+1)(n+1)(n+4)/4 =n(n+1)²(n+4)/4

1×2×3+2×3×4+3×4×5+......+n(n+1)(n+2) =1/4【1×2×3×4-0×1×2×3】+1/4【2×3×4×5-1×2×3×4】+1/4【3×4×5×6-2×3×4×5】+......+ 1/4【n(n+1)(n+2)（n+3)-(n-1)n(n+1)(n+2)】 =1/4n(n+1)(n+2)(n+3)