一道数学题

显示全部楼层 · 2012-12-9 23:10:29

解：因为1+2+……+n=n(n+1)/2所以1/(1+2+……+n)=2/n(n+1)=2*[1/n－1/(n+1)]所以1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+100)=2*[(1/1-1/2)+(1/2-1/3)+……+(1/100-1/101)]=2*(1/1-1/101)=200/101...

千问 · 2012-12-9 23:10:29

1+2+...+n=n(n+1)/2 1+1/（1+2）+1/（1+2+3）+...+1/（1+2+...+n） = 2/[1*2] + 2/[2*3] + 2/[3*4] + ... + 2/[n(n+1)] = (2/1-2/2) + (2/2-2/3) + (2/3-2/4) + ... + [2/n-2/(n+1)] = 2-2/...

千问 · 2012-12-9 23:10:29

1/[n*(n+1)/2] = 2*[1/n - 1/(n+1)]原式 =2*(1-1/2 + 1/2 - 1/3 + ... + 1/100 - 1/101) = 200/101...

千问 · 2012-12-9 23:10:29

1/(1+2+3+.....n)=2/(n+1)n=2[1/n - 1/(n+1)] 200/101...

千问 · 2012-12-9 23:10:29

1+2+...+n=n(n+1)/2原式=1+2/（2x3）+2/(3x4)+.....2/(nx(n+1))=200/101...