观察规律 1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4 求和1/1×2+1/2×3+1/3×4+.+1/2009×2010“/”仅代表分数线
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![观察规律 1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4 求和1/1×2+1/2×3+1/3×4+.+1/2009×2010“/”仅代表分数线](/uploads/image/z/5522379-51-9.jpg?t=%E8%A7%82%E5%AF%9F%E8%A7%84%E5%BE%8B+1%2F1%C3%972%3D1-1%2F2%2C1%2F2%C3%973%3D1%2F2-1%2F3%2C1%2F3%C3%974%3D1%2F3-1%2F4+%E6%B1%82%E5%92%8C1%2F1%C3%972%2B1%2F2%C3%973%2B1%2F3%C3%974%2B.%2B1%2F2009%C3%972010%E2%80%9C%2F%E2%80%9D%E4%BB%85%E4%BB%A3%E8%A1%A8%E5%88%86%E6%95%B0%E7%BA%BF)
观察规律 1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4 求和1/1×2+1/2×3+1/3×4+.+1/2009×2010“/”仅代表分数线
观察规律 1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4 求和1/1×2+1/2×3+1/3×4+.+1/2009×2010
“/”仅代表分数线
观察规律 1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4 求和1/1×2+1/2×3+1/3×4+.+1/2009×2010“/”仅代表分数线
(1)
由
1/(1×2)=(1/1)-(1/2);
1/(2×3)=(1/2)-(1/3);
1/(3×4)=(1/3)-(1/4);
从上可以看出,等式左边可以拆成二个分母组成的分式之差,分子都为1,分母分别为为n和n+1
1/[n(n+1)]=(1/n)-[1/(n+1)]
(2)证明:
等式右边=(1/n)-[1/(n+1)]
=(n+1)/[n(n+1)]-n/[n(n+1)]
=(n+1-n)/[n(n+1)]
=1/[n(n+1)]
=左边
所以等式成立
(3)求和:观察后可以发现好多项可以相互抵消
1/(1×2)+1/(2×3)+1/(3×4)+……+1/(2009×2010)
=1-1/2+1/2-1/3+1/3-1/4+-------+1/2008-1/2009+1/2009-1/2010
=1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+-------+1/2008+(-1/2009+1/2009)-1/2010
=1-1/2010
=2009/2010
分数裂项题:
1/1×2+1/2×3+1/3×4+....+1/2009×2010
=1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010
=1-1/2010
=2009/2010
将1/1×2=1-1/2,1/2×3=1/2-1/3,1/3×4=1/3-1/4排成一列相加可知,剩余第一项跟最后一项,所以
1/1×2+1/2×3+1/3×4+....+1/2009×2010=1-1/2010=2009/2010