观察下面的变形规律1/(1×2)=1-1/2;1/(2×3)=1/2-1/3;1/(3×4)=1/3-1/4;.(1)若n为正整数,请你猜想1/[n(n+1)]=?(2)证明你猜想的结论;(3)求和:1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2009×2010)·
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![观察下面的变形规律1/(1×2)=1-1/2;1/(2×3)=1/2-1/3;1/(3×4)=1/3-1/4;.(1)若n为正整数,请你猜想1/[n(n+1)]=?(2)证明你猜想的结论;(3)求和:1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2009×2010)·](/uploads/image/z/246155-59-5.jpg?t=%E8%A7%82%E5%AF%9F%E4%B8%8B%E9%9D%A2%E7%9A%84%E5%8F%98%E5%BD%A2%E8%A7%84%E5%BE%8B1%2F%281%C3%972%29%3D1-1%2F2%EF%BC%9B1%2F%282%C3%973%29%3D1%2F2-1%2F3%EF%BC%9B1%2F%283%C3%974%29%3D1%2F3-1%2F4%EF%BC%9B.%EF%BC%881%EF%BC%89%E8%8B%A5n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E8%AF%B7%E4%BD%A0%E7%8C%9C%E6%83%B31%2F%5Bn%28n%2B1%29%5D%3D%3F%EF%BC%882%EF%BC%89%E8%AF%81%E6%98%8E%E4%BD%A0%E7%8C%9C%E6%83%B3%E7%9A%84%E7%BB%93%E8%AE%BA%EF%BC%9B%EF%BC%883%EF%BC%89%E6%B1%82%E5%92%8C%EF%BC%9A1%2F%281%C3%972%29%2B1%2F%282%C3%973%29%2B1%2F%283%C3%974%29%2B.%2B1%2F%282009%C3%972010%29%C2%B7)
观察下面的变形规律1/(1×2)=1-1/2;1/(2×3)=1/2-1/3;1/(3×4)=1/3-1/4;.(1)若n为正整数,请你猜想1/[n(n+1)]=?(2)证明你猜想的结论;(3)求和:1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2009×2010)·
观察下面的变形规律1/(1×2)=1-1/2;1/(2×3)=1/2-1/3;1/(3×4)=1/3-1/4;.
(1)若n为正整数,请你猜想1/[n(n+1)]=?
(2)证明你猜想的结论;
(3)求和:1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2009×2010)·
观察下面的变形规律1/(1×2)=1-1/2;1/(2×3)=1/2-1/3;1/(3×4)=1/3-1/4;.(1)若n为正整数,请你猜想1/[n(n+1)]=?(2)证明你猜想的结论;(3)求和:1/(1×2)+1/(2×3)+1/(3×4)+.+1/(2009×2010)·
(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
(1)若n为正整数,1/[n(n+1)]=1/n-1/(n+1)
(2)右边=1/n-1/(n+1)=(n+1)/n(n+1)-n(n(n+1)=1/n(n+1)=左边
(3)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-/2010
=2009/2010
1.=1/n-1/(n+1)2.因1/(1×2)=1-1/2;1/(2×3)=1/2-1/3;1/(3×4)=1/3-1/4所以1/n-1/(n+1)
3.=1-1/2+1/2-1/3……+1/2009-1/2010
=1-1/2010
=2009/2010
1/n-1/n+1
123=1/2×456=1/3×789
(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/(...
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(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
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