（1／1^2+2）+（1／2^2+4）+...+（1／n^2+2n)
=(1/2)[(1/1)-(1/3)]+(1/2)[(1/2)-(1/4)]+(1/2)[(1/3)-(1/5)]+...
...+(1/2)[1/(n-1)-1/(n+1)]+(1/2)[1/n-1/(n+2)]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-(2n+3)/(n+1)(n+2)]
求数列（1／1^2+2）+（1／2^2+4）+...+（1／n^2+2n)的前n项和
（1／1^2+2）+（1／2^2+4）+...+（1／n^2+2n)
=(1/2)[(1/1)-(1/3)]+(1/2)[(1/2)-(1/4)]+(1/2)[(1/3)-(1/5)]+...
...+(1/2)[1/(n-1)-1/(n+1)]+(1/2)[1/n-1/(n+2)]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-(2n+3)/(n+1)(n+2)]