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1/1+2+1/1+2+3+...+1/1+2+3+...+49各位高手帮帮忙
题目详情
1/1+2+1/1+2+3+...+1/1+2+3+...+49
各位高手帮帮忙
各位高手帮帮忙
▼优质解答
答案和解析
1+2+3+……+n=n(n+1)/2
所以,1/(1+2+3+……+n)=2/n(n+1)=2*[(1/n)-(1/n+1)]
所以原式=2*[(1/2)-(1/3)]+2*[(1/3)-(1/4)]+……+2*[(1/49)-(1/50)]
=2*[(1/2)-(1/3)+(1/3)-(1/4)+……+(1/49)-(1/50)]
=2*[(1/2)-(1/50)]
=1-(1/25)
=24/25.
所以,1/(1+2+3+……+n)=2/n(n+1)=2*[(1/n)-(1/n+1)]
所以原式=2*[(1/2)-(1/3)]+2*[(1/3)-(1/4)]+……+2*[(1/49)-(1/50)]
=2*[(1/2)-(1/3)+(1/3)-(1/4)+……+(1/49)-(1/50)]
=2*[(1/2)-(1/50)]
=1-(1/25)
=24/25.
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