探究性问题:11×2=11−12,12×3=12−13,13×4=13−14,则1n(n+1)=1n-1n+11n-1n
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探究性问题:
=
−
1 |
1×2 |
1 |
1 |
1 |
2 |
![探究性问题:11×2=11−12,12×3=12−13,13×4=13−14,则1n(n+1)=1n-1n+11n-1n](/uploads/image/z/19390328-8-8.jpg?t=%E6%8E%A2%E7%A9%B6%E6%80%A7%E9%97%AE%E9%A2%98%EF%BC%9A11%C3%972%EF%BC%9D11%E2%88%9212%EF%BC%8C12%C3%973%EF%BC%9D12%E2%88%9213%EF%BC%8C13%C3%974%EF%BC%9D13%E2%88%9214%EF%BC%8C%E5%88%991n%28n%2B1%29%3D1n-1n%2B11n-1n)
根据已知的三个等式,总结规律得
1
n(n+1)=
1
n-
1
n+1,
(1)原式=
1
(x+1)(x+2)+
1
(x+2)(x+3)+
1
(x+3)(x+4)
=
1
x+1-
1
x+2+
1
x+2-
1
x+3+
1
x+3-
1
x+4=
1
x+1-
1
x+4=
3
(x+1)(x+4);
(2)由
a−1+(ab−2)2=0得:a-1=0且ab-2=0,
解得a=1且ab=2,
所以b=2,
则原式=
1
ab+
1
(a+1)(b+1)+…+
1
(a+2010)(b+2010),
=
1
1×2+
1
2×3+…+
1
2011×2012,
=1-
1
2+
1
2-
1
3+
1
3-
1
4+…+
1
2010-
1
2011+
1
2011-
1
2012=1-
1
2012=
2011
2012.
故答案为:
1
n-
1
n+1.
1
n(n+1)=
1
n-
1
n+1,
(1)原式=
1
(x+1)(x+2)+
1
(x+2)(x+3)+
1
(x+3)(x+4)
=
1
x+1-
1
x+2+
1
x+2-
1
x+3+
1
x+3-
1
x+4=
1
x+1-
1
x+4=
3
(x+1)(x+4);
(2)由
a−1+(ab−2)2=0得:a-1=0且ab-2=0,
解得a=1且ab=2,
所以b=2,
则原式=
1
ab+
1
(a+1)(b+1)+…+
1
(a+2010)(b+2010),
=
1
1×2+
1
2×3+…+
1
2011×2012,
=1-
1
2+
1
2-
1
3+
1
3-
1
4+…+
1
2010-
1
2011+
1
2011-
1
2012=1-
1
2012=
2011
2012.
故答案为:
1
n-
1
n+1.
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