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化簡(jiǎn),前后項(xiàng)可以抵消,就剩下首尾項(xiàng),即[(99*100*101*102)-(0*1*2*3)]/4=25497450下面一題,題目中多打了一個(gè)91...能不能寫的再簡(jiǎn)便一點(diǎn)?哪一個(gè)？上面一個(gè)還是下面一個(gè)？第二題是多打個(gè)91，抱歉。請(qǐng)把全部計(jì)算過(guò)程寫下來(lái)，謝謝！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（上式整體除以3）兩式相減得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（上式整體乘以3）兩式相減得3B-B=1-1/243=242/243=2BB=121/243整體A+B=364+121/243清楚嗎？想寫的簡(jiǎn)便的話可以直接寫原式=(243-1/3)/(1-1/3)+(1-1/243)/(3-1)=364+121/243不過(guò)這樣思路有點(diǎn)跳躍第一題1×2×3+2×3×4+3×4×5...+99×101====第二題243+81+27+9+3+1+1/3+1/9+1/27+1/81+1/243===第一題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第二題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