Question

计算1×2×3＋2×3×4＋3×4×5＋……. ＋100×101×102？

Answer 1

数列：an=n(n+1)(n+2)=n³+3n²+2n
原式=(1³+2³+···+100³)+3(1²+2²+···+100²)+2(1+2+···+100)

1³+2³+···+n³ =[n(n+1)]²/4
1²+2²+···+n²=[n(n+1)(2n+1)]/6
1+2+···+n=n(n+1)/2
(1³+2³+···+n³)+3(1²+2²+···+n²)+2(1+2+···+n)
=[n(n+1)]²/4+[n(n+1)(2n+1)]/2+n(n+1)
=n(n+1){[n(n+1)]/4+(2n+1)/2+1}
=1/4n(n+1)(n²+n+4n+2+4)
=1/4n(n+1)(n²+5n+6)
=1/4n(n+1)(n+2)(n+3)
原式=1/4x100x101x102x103=106 215 660/4=24553915

Answer 2

1×2×3+2×3×4+3×4×5+···+100×101×102
= 2×(2-1)×(2+1)+3×(3-1)×(3+1)+4×(4-1)×(4+1)+···+101×(101-1)×(101+1)
= 2×(2²-1²)+3×(3²-1²)+4×(4²-1²)+···+101×(101²-1²)
= 2³-2+3³-3+4³-4+···+101³-101
= (1³+2³+3³+4³+···+101³-1)-(1+2+3+4+···+101-1)
= {[101×(101+1)÷2]²-1}-[101×(101+1)÷2-1]
= 26532800-5150
= 26527650