①计算[1/(1×2)]+[1/(2×3)]+[1/(3×4)]+……+[1/(10×11)].
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①计算[1/(1×2)]+[1/(2×3)]+[1/(3×4)]+……+[1/(10×11)].
②利用上题的思路计算:{1/[a(a+1)]}+{1/[(a+1)(a+2)]}+{1/[(a+2)(a+3)]}+……+{1/[(a+2008)(a+2009)]}+{1/[(a+2009)(a+2010)]}的值.
②利用上题的思路计算:{1/[a(a+1)]}+{1/[(a+1)(a+2)]}+{1/[(a+2)(a+3)]}+……+{1/[(a+2008)(a+2009)]}+{1/[(a+2009)(a+2010)]}的值.
[1/(1×2)]+[1/(2×3)]+[1/(3×4)]+……+[1/(10×11)].
=1-1/2+1/2-1/3+1/3-1/4+……+1/10+1/11
=1-1/11
=10/11
{1/[a(a+1)]}+{1/[(a+1)(a+2)]}+{1/[(a+2)(a+3)]}+……+{1/[(a+2008)(a+2009)]}+{1/[(a+2009)(a+2010)]}
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+……+1/(a+2009)-1/(a+2010)
=1/a-1/(a+2010)
=2010/[a(a+2010)]
=1-1/2+1/2-1/3+1/3-1/4+……+1/10+1/11
=1-1/11
=10/11
{1/[a(a+1)]}+{1/[(a+1)(a+2)]}+{1/[(a+2)(a+3)]}+……+{1/[(a+2008)(a+2009)]}+{1/[(a+2009)(a+2010)]}
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+……+1/(a+2009)-1/(a+2010)
=1/a-1/(a+2010)
=2010/[a(a+2010)]
①计算[1/(1×2)]+[1/(2×3)]+[1/(3×4)]+……+[1/(10×11)].
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