求数列之和 1/（1*3）+1/（2*4）+1/（3*5）+.+1/n*（n+2）之和、 怎么求,
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1/(1×3)+1/(2×4)+1/(3×5)+…+1/[n(n+2)]
=1/2(1-1/3)+1/2(1/2-1/4)+1/2(1/3-1/5)+…+1/2[1/n-1/(n+2)]
=1/2[1-1/3+1/2-1/4+1/3-1/5+…+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
=3/4-1/[2(n+1)]-1/[2(n+2)]
1/(1×3)+1/(2×4)+1/(3×5)+…+1/[n(n+2)]
=1/2(1-1/3)+1/2(1/2-1/4)+1/2(1/3-1/5)+…+1/2[1/n-1/(n+2)]
=1/2[1-1/3+1/2-1/4+1/3-1/5+…+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
=3/4-1/[2(n+1)]-1/[2(n+2)]
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