Question

1×2+2×3+3×4...+n(n+1)= 1×2×3＋2×3×4＋3×4×5...＋7×8×9＝

Answer 1

1×2+2×3+3×4+...+n(n+1)
=1×(1+1)+2×(2+1)+3×(3+1)+...+n(n+1)
=1²+2²+3²+...+n²+1+2+3+...+n
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(2n+1+3)/6
=2n(n+1)(n+2)/6
=n(n+1)(n+2)/3

1×2×3+2×3×4+...+7×8×9
=1×(1+1)(1+2)+2×(2+1)(2+2)+...+7×(7+1)(7+2)
=1³+3×1²+2×1+2³+3×2²+2×2+...+7³+3×7²+2×7
=(1³+2³+...+7³)+3(1²+2²+...+7²)+2(1+2+...+7)
=[(7×8)/2]²+3×7×8×15/6+7×8
=784+420+56
=1260

Answer 2

1.
因 n(n+1) = n^2 + n
则
1*2+2*3+3*4......+n*(n+1)

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

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

= 1/3n(n+1)(n+2)
2.
观察知每项都是（n+1）^3-n,
1*2*3+2*3*4+3*4*5+···+7*8*9
= (2³ - 2) + (3³ - 3) + …… + (8³ - 8)
= 1³ + 2³ + 3³ + …… + 8³ - (1+2+3+……+8)
套用连续立方和公式、等差数列求和公式
= [8 * (8+ 1)/2]^2 - (1+8) * 8 / 2
=1296 - 36
= 1260

Answer 3

1260