·请教·一道关于数列的题

显示全部楼层 · 2009-8-18 16:09:36

1+2+……+n=n(n+1)/2所以1/(1+2+……+n)=2/n(n+1)=2*[1/n-1/(n+1)]所以原式=2*(1-1/2)+2*(1/2-1/3)+……+2*[1/n-1/(n+1)]=2*[1-1/2+1/2-1/3+……+1/n-1/(n+1)]=2*[1-1/(n+1)]=2n/(n+1)...

千问 · 2009-8-18 16:09:36

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)]所以前n项和=2*(1-1/2)+2*(1/2-1/3)+……+2*[1/n-1/(n+1)]=2*[1-1/2+1/2-1/3+……+1/n-1/(n+1)]=2*[1-1/(n+1)]=2n/(n+1)...

千问 · 2009-8-18 16:09:36

前n项和=1+2/6+2/12+......+2/((n*(n+1))=1+2(1/2-1/3+1/3-1/4+.....-1/n+1/n-1/(n+1))=2*n/(n+1)...

千问 · 2009-8-18 16:09:36

裂项法Sn=1+1-1/2+1/2-1/3+…1/(n-1)-1/n=1+1-1/n=1+(n-1)/n...