4⼀1*2*3+5⼀2*3*4+6⼀3*4*5+...+11⼀8*9*10

从8个分数的特点可以写出 (n+3)/n*(n+1)*(n+2)=x/n*(n+1)-y/(n+1)*(n+2),
对右边通分 [x(n+2)-ny]=n*(n+1)*(n+2)=(n+3)/n*(n+1)*(n+2)
两边的分子相等 (x-y)n+2x=n+3
两边的同类项系数对应相等x-y=1, 2x=3, 解得x=3/2, y=1/2

原式=1/2{[3/1*2-1/2*3]+[3/2*3-1/3*4]+[3/3*4-1/4*5]+[3/4*5-1/5*6]+[3/5*6-1/6*7]+[3/6*7-1/7*8]+[3/7*8-1/8*9]+[3/8*9-1/9*10]
每一个中括号内有二个数，把前一个括号内的第二数与后一个括号内的第一个数相加
=1/2{3/2+(1/3+1/6)+(1/10+1/15)+(1/21+1/28)+1/36-1/90}
=1/2{(3/2+1/2)+(1/6+1/12)+(1/36-1/90)}
=1/2{2+1/4+1/60}
=17/15

考察一般项第k项：
(k+3)/[k(k+1)(k+2)]
=[(k+1)+(k+2)-k]/[k(k+1)(k+2)]
=1/[k(k+2)+1/[k(k+1)]-1/[(k+1)(k+2)]
=(1/2)[1/k-1/(k+2)]+1/k-1/(k+1)-[1/(k+1)-1/(k+2)]
=(3/2)/k+(1/2)/(k+2)]-2/(k+1)
=(3/2)[1/k-1/(k+1)]-(1/2)[1/(k+1)-1/(k+2)]

4/(1×2×3)+5/(2×3×4)+6/(3×4×5)+...+11/(8×9×10)
=(3/2)(1/1-1/2+1/2-1/3+...+1/8-1/9)-(1/2)(1/2-1/3+1/3-1/4+...+1/9-1/10)
=(3/2)(1-1/9)-(1/2)(1/2-1/10)
=4/3-1/5
=17/15

知识推广：
4/(1×2×3)+5/(2×3×4)+...+(n+3)/[n(n+1)(n+2)]
=(3/2)[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]-(1/2)[1/2-1/3+1/3-1/4+...+1/(n+1)-1/(n+2)]
=(3/2)[1-1/(n+1)]-(1/2)[1/2-1/(n+2)]
=3n/[2(n+1)]-n/[4(n+2)]
=n(5n+11)/[4(n+1)(n+2)]
推导的这个结果可以求任意项的和，不局限于本题的n=8，可以计算n很大时的值。