Question

计算1*2*3+2*3*4+3*4*5+4*5*6+......+n(n+1)(n+2)

1^3+2^3+……+n^3=[n(n+1)/2]^2
1^2+2^2+……+n^2=n(n+1)(2n+1)/6
是怎么算的啊？

Answer 1

n(n+1)(n+2)=n^3+3n^2+2n
1×2×3+2×3×4+3×4×5+……n(n+1)(n+2)
=(1^3+2^3+……+n^3)+3*(1^2+2^2+……+n^2)+2*(1+2+……+n)
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+……+n=n(n+1)/2

所以原式=[n(n+1)/2]^2+3n(n+1)(2n+1)/6+2n(n+1)/2
=n^2(n+1)^2/4+n(n+1)(2n+1)/2+n(n+1)
=[n(n+1)/4]*[n(n+1)+2(2n+1)+4]
=[n(n+1)/4]*(n^2+5n+6)
=n(n+1)(n+2)(n+3)/4

Answer 2

原式中第i项 为i(i+1)(1+2)=(i+1)^3-(i+1)
原式=(2^3-2)+(3^3-3)+(4^3-4)+……+(n^3-n)+((n+1)^3-(n+1))
=(2^3+3^3+……+(n+1)^3)-(2+3+……+n+(n+1))

因为1^3+2^3+...+n^3=[n(n+1)/2]^2
原式={[(n+1)(n+2)/2]^2-1}-(n+1)(n+2)/2-1
=[(n+1)(n+2)/2]^2-(n+1)(n+2)/2-2

Answer 3

这种题直接代公式就可以了 1*2*3+2*3*4+3*4*5+4*5*6+......+n(n+1)(n+2)
=n(n+1)(n+2)(n+3)/4=
4
6*C
n+3

表示从n+3个数中选4个 再乘以6

Answer 4

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