1/6+1/12+1/20+...+1/(n^2+n)等于?
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1/6+1/12+1/20+...+1/(n^2+n)等于?1/6+1/12+1/20+...+1/(n^2+n)等于?1/6+1/12+1/20+...+1/(n^2+n)等于?1/(n^2+n)=
1/6+1/12+1/20+...+1/(n^2+n)等于?
1/6+1/12+1/20+...+1/(n^2+n)等于?
1/6+1/12+1/20+...+1/(n^2+n)等于?
1/(n^2+n)=1/n(n+1)=1/n-1/(n+1)
所以
1/6+1/12+1/20+...+1/(n^2+n）
=1/2-1/3+1/3-1/4+1/4-1/5+.+1/n-1/(n+1)
=1/2-1/(n+1)
=(n-1)/(2n+2)
原式=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...+[1/n-1/(n+1)]
=1/2-1/(n+1)
=(n-1)/(2n+2)
=1/2-1/3+1/3-1/4+...1/n-1/(n+1)
=1/2-1/(n+1)
=(n-1)/2(n+1)