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...能不能写的再简便一点?哪一个？上面一个还是下面一个？第二题是多打个91，抱歉。请把全部计算过程写下来，谢谢！243+81+27+9+3+1+1/3+1/9+1/27+1/81+1/243A=243+81+27+9+3+1A/3=81+27+9+3+1+1/3（上式整体除以3）两式相减得A-A/3=243-1/3=242又2/3=2A/3 A=(242又2/3)*3/2=364B=1/3+1/9+1/27+1/81+1/2433B=1+1/3+1/9+1/27+1/81（上式整体乘以3）两式相减得3B-B=1-1/243=242/243=2BB=121/243整体A+B=364+121/243清楚吗？想写的简便的话可以直接写原式=(243-1/3)/(1-1/3)+(1-1/243)/(3-1)=364+121/243不过这样思路有点跳跃第一题1×2×3+2×3×4+3×4×5...+99×101====第二题243+81+27+9+3+1+1/3+1/9+1/27+1/81+1/243===第一题1×2×3+2×3×4+3×4×5...+99×101=[(1*2*3*4)-(1*2*3*0)]/4+[(2*3*4*5)-(2*3*4*1)]/(5-1)+……+[(99*100*101*102)-(99*100*101*98)]/(102-98)=[(99*100*101*102)-(0*1*2*3)]/4=25497450第二题243+81+27+9+3+1+1/3+1/9+1/27+1/81+1/243=[(243+81+27+9+3+1)-(243+81+27+9+3+1)/3]/(1-1/3)+[(1/3+1/9+1/27+1/81+1/243)*3-(1/3+1/9+1/27+1/81+1/243)]/(3-1)=(243-1/3)/(1-1/3)+(1-1/243)/(3-1)=364+121/243