1×2×3+2×3×4+3×4×5...+99×101

n(n+1)(n+2)=[n(n+1)(n+2)(n+3)-n(n+1)(n+2)(n-1)]/[(n+3)-(n-1)]
=[n(n+1)(n+2)(n+3)-n(n+1)(n+2)(n-1)]/4
化简，前后项可以抵消，就剩下首尾项，即[(99*100*101*102)-(0*1*2*3)]/4=25497450

下面一题，题目中多打了一个91吧？在27和9中间，不和规律的
前半段A=243+81+27+9+3+1

A/3=81+27+9+3+1+1/3

A-A/3=243-1/3=242又2/3=2A/3
A=(242又2/3)*3/2=364
后半段B=1/3+1/9+1/27+1/81+1/243

3B=1+1/3+1/9+1/27+1/81
3B-B=1-1/243=242/243=2B
B=121/243
整体A+B=364+121/243

若要加上91，则答案为455+121/243

(n-1)×n×(n+1)=(n+1)³-(n+1)
1×2×3+2×3×4+3×4×5+……+99×100*101=(2³-2)+(3³-3)+(4³-4)+……+(100³-100)
=(1³+2³+3³+4³+……+100³)-(1+2+3+4+……+100)
=100²(100+1)²/4-100*(100+1)/2
=5050*5049=25497450；
243+81+27+91+9+3+1+(1/3)+(1/9)+(1/27)+(1/81)+(1/243)
=(243+81+27+9+3+1)+91+[(1/3)+(1/9)+(1/27)+(1/81)+(1/243)]
=[243*3-1]/(3-1) +91 +[(1/3)-(1/243)*(1/3)]/(1-1/3)
=364+91+(121/243)=455+(121/243)；