考察一般项：
1/(n²+2n)=1/[n(n+2)]=(1/2)[1/n -1/(n+2)]
1/(1²+2)+1/(2²+4)+...+1/(n²+2n)
=(1/2)[1/1-1/3+1/2-1/4+...+1/n-1/(n+2)]
=(1/2)[(1/1+1/2+...+1/n)-(1/3+1/4+...+1/(n+1)+1/(n+2)]
=(1/2)[1+1/2 -1/(n+1)-1/(n+2)]
=3/4 -1/[2(n+1)]- 1/[2(n+2)]
1²+2分之一+2²+4分之一+3²+6分之一+……+n²+2n分之一
=[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-(2n+3)/[2(n+1)(n+2)]
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