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Algebra Difficulty 5.5 AIME, harder Prove it Ukraine

Find distinct numbers aa, bb, cc, dd such that:
12011=aa+1+bb+1−cc+1−dd+1 \frac{1}{2011} = \frac{a}{a+1} + \frac{b}{b+1} - \frac{c}{c+1} - \frac{d}{d+1}

Solution

We use the following formulas:
1n=n−1n(n−1)=1n−1−1n(n−1),1n=n+1n(n+1)=1n+1+1n(n+1) \frac{1}{n} = \frac{n-1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n(n-1)}, \quad \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{1}{n+1} + \frac{1}{n(n+1)}
Using them we arrive at:
12011=12010−12010⋅2011=12011+12011⋅2012−12010⋅2011+1−12010⋅2011(2010⋅2011+1)=1a+1+1b+1−1c+1−1d+1 \frac{1}{2011} = \frac{1}{2010} - \frac{1}{2010 \cdot 2011} = \frac{1}{2011} + \frac{1}{2011 \cdot 2012} - \frac{1}{2010 \cdot 2011 + 1} \\ - \frac{1}{2010 \cdot 2011(2010 \cdot 2011+1)} = \frac{1}{a+1} + \frac{1}{b+1} - \frac{1}{c+1} - \frac{1}{d+1}
We add 1 to the first two fractions and subtract one from the last two, we get the following representation:
12011=2010⋅2011⋅2012−12011⋅2011⋅2012+2010⋅20112010⋅2011+1++2010⋅2011(2010⋅2011+1)−12010⋅2011(2010⋅2011+1)=aa+1+bb+1−cc+1−dd+1 \frac{1}{2011} = \frac{2010 \cdot 2011 \cdot 2012 - 1}{2011 \cdot 2011 \cdot 2012} + \frac{2010 \cdot 2011}{2010 \cdot 2011 + 1} + \\ + \frac{2010 \cdot 2011(2010 \cdot 2011 + 1) - 1}{2010 \cdot 2011(2010 \cdot 2011 + 1)} = \frac{a}{a+1} + \frac{b}{b+1} - \frac{c}{c+1} - \frac{d}{d+1}

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