计算：1/1×2×3×4＋1/2×3×4×5＋……＋1/17×18×19×20求详细过程,谢谢!

1/[n*(n+1)*(n+2)*(n+3)]=1/[(n^2+3*n)*(n^2+3n+2)]
=1/2 *[1/(n^2+3n)-1/(n^2+3n+2)]
=1/6 *[1/n-1/(n+3)]-1/2 *[1/(n+1)-1/(n+2)]
所以原式=1/6 *(1+1/2+……+1/17-1/4-1/5-……-1/20)-1/2 *(1/2+1/3+……+1/18-1/3-1/4-……1/19)
=1/6 *(1+1/2+1/3-1/18-1/19-1/20)-1/2 *(1/2-1/19)
=5729/20520-17/76
=1139/20520

1/[n(n+1)(n+2)(n+3)]=1/{[n(n+3)]*[(n+1)(n+2)]}=1/2*{1/[n(n+3)]-1/[(n+1)(n+2)]}
1/(1*2*3*4)+1/(2*3*4*5)+...+1/(17*18*19*20)
=1/2*[1/(1*4)-1/(2*3)+1/(2*5)-1/(3*4)+...+1/(17*20)-1/(18*19)]
=1/2*[1/(1*4)+1/(2*5)+1/(3*6)+...+1/(17*20)]-[1/(2*3)+1/(3*4)+1/(4*5)+...+1/(18*19)]
因为1/[a*(a+3)]=1/3*[1/a-1/(a+3)]
所以1/(1*4)+1/(2*5)+1/(3*6)+...+1/(17*20)=1/1-1/4+1/2-1/5+1/3-1/6+...+1/17-1/20=1+1/2+1/3-1/18-1/19-1/20
因为1/[b*(b+1)]=1/b-1/(b+1)
所以1/(2*3)+1/(3*4)+1/(4*5)+...+1/(18*19)=1/2-1/3+1/3-1/4+...+1/18-1/19=1/2-1/19
原式=1/2*[(1+1/2+1/3-1/18-1/19-1/20)-(1/2-1/19)]=1/2*(1+1/3-1/18-1/20)=1/2*(180+60-10-9)/180=221/360

1/1*2*3 -1/2*3*4 =(4 -1)/1*2*3*4 = 3/1*2*3*4,同理 3/2*3*4*5 =1/2*3*4 -1/3*4*5,所以 1/1*2*3*4 +1/2*3*4*5 +------+1/17*18*19*20=（1/3）*（3/1*2*3*4 +3/2*3*4*5 +------3/17*18*19*20）= （1/3）*（1/1*2*3 -...

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