求1⼀(1*2*3*4)+1⼀(2*3*4*5)+1⼀(3*4*5*6)至n项之和

先看 1/(n*(n+1)*(n+2)*(n+3) )
=1/(n^2+3n)*(n^2+3n+2)
=1/2(1/(n^2+3n)-1/(n^2+3n+2))
=1/2(1/3(1/n-1/(n+3))-(1/(n+1)-1/(n+2))
=1/6 (1/n-1/(n+3)) -1/2(1/(n+1)-1/(n+2))
所以1/(1*2*3*4)=1/6 (1-1/4) -1/2(1/2-1/3)
1/(2*3*4*5)=1/6(1/2-1/5)-1/2(1/3-1/4)
1/(3*4*5*6)=1/6(1/3-1/6)-1/2(1/4-1/5)
1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)+...+1/(n*(n+1)*(n+2)*(n+3))
=1/6(1-1/4+1/2-1/5+1/3-1/6+......+1/n-1/(n+3))-1/2(1/2-1/3+1/3-1/4+1/4-1/5+....+1/(n+1)-1/(n+2))
=1/6(1-1/4+1/2-1/5+1/3-1/6+...+1/n-1/(n+3))-1/2(1/2-1/(n+2))
=1/6(1+1/2+1/3 -1/(n+1)-1/(n+2)-1/(n+3)) -1/2(1/2-1/(n+2))