作业帮1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.1/(1+2+3+4+5+6+.+50)=?
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1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.1/(1+2+3+4+5+6+.+50)=?
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.1/(1+2+3+4+5+6+.+50)=?
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.1/(1+2+3+4+5+6+.+50)=?
1/(1+2+3+……+n)=1/[n(n+1)/2]=2/[n(n+1)]
=2*[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.1/(1+2+3+4+5+6+.+50)
=2*[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/50-1/51)]
=2*(1-1/51)
=100/51