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an=2n,
bn=2/[(n+3)(an+2)]=2/[(n+3)(2n+2)]=1/[(n+3)(n+1)]=(1/2)[1/(n+1)-1/(n+3)],
求和：
Sn=b1+b2+b3+...+b(n-1)+bn
=(1/2)×{(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+...+[1/n-1/(n+2)]+[1/(n+1)-1/(n+3)]}
=(1/2)×[1/2+1/3-1/(n+2)-1/(n+3)]
=(1/2)×{5/6-1/[(n+2)(n+3)]}
=5/12-1/[2(n+2)(n+3)]
bn=2/[(n+3)(an+2)]=2/[(n+3)(2n+2)]=1/[(n+3)(n+1)]=(1/2)[1/(n+1)-1/(n+3)],
求和：
Sn=b1+b2+b3+...+b(n-1)+bn
=(1/2)×{(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+...+[1/n-1/(n+2)]+[1/(n+1)-1/(n+3)]}
=(1/2)×[1/2+1/3-1/(n+2)-1/(n+3)]
=(1/2)×{5/6-1/[(n+2)(n+3)]}
=5/12-1/[2(n+2)(n+3)]
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