已知ak=(k+2)/k!+(k+1)!+(k+2)!(k>=1),则其前100项的和是?

Question

已知ak=(k+2)/k!+(k+1)!+(k+2)!(k>=1),则其前100项的和是?

Answer 1

k!+(k+1)!+(k+2)!
=k![1+(k+1)+(k+1)(k+2)]
=k!(k^2+4k+4)
=k!(k+2)^2
所以ak=(k+2)/[k!(k+2)^2]=1/[k!(k+2)]=(k+1)/k!(k+1)(k+2)]=(k+1)/(k+2)!=[(k+2)-1]/(k+2)!=(k+2)/(k+2)!-1/(k+2)!
=1/(k+1)!-1/(k+2)!
所以S100=(1/2!-1/3!)+(1/3!-1/4!)+……+(1/101!-1/102!)
=1/2-1/102!