Question

1×2×3＋2×3×4＋3×4×5+.......n×（n+1）×（n+2）

需要详细的解析过程

Answer 1

原式
= (2-1)*2(2+1) + (3-1)*3*(3+1) + …… + (N+1-1)×(n+1)×(N+1+1)
= 2*(2² - 1) + 3(3² - 1) + …… + (N+1)*[(N+1)² - 1]
= 2³ - 2 + 3³ - 3 + …… + (N+1)³ - (N+1)
= 1³ + 2³ + 3³ + …… + (N+1)³ - [1 + 2 + 3 + …… + (N+1)]
= [(1+2+……+(n+1) ]^2 - (N+2)(N+1)/2
= (N+2)²(N+1)²/4 - (N+2)(N+1)/2
=(N^4 + 6N³ + 11N² + 6N )/4

其中用到了立方和公式

Answer 2

1^3+2^3+3^3+4^3+……+n^3=(n+1)^2*n^2/4

n(n+1)(n+2)
=n (n+2)(n+1)
=(n^2+2n)(n+1)
=(n^2+2n+1-1)(n+1)
=[(n+1)^2-1](n+1)
=(n+1)^3-n-1

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

Answer 3

1×2×3＋2×3×4＋3×4×5+.......n×（n+1）×（n+2）
=(N^4 + 6N³ + 11N² + 6N )/4
=(n+1)^2*(n+2)^2/4-n^2-n-1 一样吧