通项为n/(n+1)(n+2)(n+3)，求前n项和

通项为n/(n+1)(n+2)(n+3)，求前n项和

sn=1/2*3*4+2/3*4*5+3/4*5*6+4/5*6*7+...+n/(n+1)(n+2)(n+3)

    =1/2*(1/2*3-1/3*4)+2/2*(1/3*4-1/4*5)+3/2*(1/4*5-1/5*6)+...+n/2[1/(n+1)(n+2)-1/(n+2)(n+3)]

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

 =1/2*1/2*3+1/2*1/3*4+1/2*1/4*5+...+1/2*1/(n+1)(n+2)-n/2*1/(n+2)(n+3)

=1/2[1/2*3+1/3*4+1/4*5+...+1/(n+1)(n+2)]-n/2*1/(n+2)(n+3)

=n(n+1)/4(n+2)(n+3)

an=n/[(n+1)(n+2)(n+3)]

=[(n+1)-1]/[(n+1)(n+2)(n+3)]

=1/[(n+2)(n+3)]-1/[(n+1)(n+2)(n+3)]

=1/(n+2)-1/(n+3)+[(n+1)(n+3)-(n+2)^2]/[(n+1)(n+2)(n+3)]

=1/(n+2)-1/(n+3)+1/(n+2)-(n+2)/[(n+1)(n+3)]

=2/(n+2)-1/(n+3)-[(n+1)+1]/[(n+1)(n+3)]

=2/(n+2)-1/(n+3)-1/(n+3)-1/[(n+1)(n+3)]

=2/(n+2)-2/(n+3)-1/2[1/(n+1)-1/(n+3)]

=2/(n+2)-5/[2(n+3)]-1/[2(n+1)]

=4/[2(n+2)]-5/[2(n+3)]-1/[2(n+1)]

故各项为

a1=4/(2*3)-5/(2*4)-1/(2*2)

a2=4/(2*4)-5/(2*5)-1/(2*3)

a3=4/(2*5)-5/(2*6)-1/(2*4)

....

an=4/[2(n+2)]-5/[2(n+3)]-1/[2(n+1)]

前n项和=

借用一下用组合数的性质： (n+3)(n+2)(n+1)n  = 4！* C（n+3,4） 则 Sn=4!*[ C(4,4)+C(5,4)+C(6,4)+……+C(n+3,4)]      =4!*C(n+4,5)      =1/5(n+4)(n+3)(n+2)(n+1)n