1\2+1\6+1\12+.+n(n+1)等于多少啊
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1\2+1\6+1\12+.+n(n+1)等于多少啊1\2+1\6+1\12+.+n(n+1)等于多少啊1\2+1\6+1\12+.+n(n+1)等于多少啊因为:1/[n(n+1)]=1/n-1/(n
1\2+1\6+1\12+.+n(n+1)等于多少啊
1\2+1\6+1\12+.+n(n+1)等于多少啊
1\2+1\6+1\12+.+n(n+1)等于多少啊
因为:1/[n(n+1)]=1/n-1/(n+1)
所以:
1/2+1/6+1/12+……+1/[n(n+1)]
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+…+[(1/n-1/(n+1)]
=1-1/2+1/2-1/3+1/3……-1/(n+1)
=1-1/(n+1)
=(n+1)/(n+1)-1/(n+1)
=n/(n+1)
有才
1/2+1/6+1/12+...+1/n(n+1)
=1/1*2+1/2*3+1/3*4+...+1/n*(n+1)
=1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
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