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Algebra Difficulty 4.0 AMC 10/12 Find the answer Saudi Arabia

Evaluate the sum
1+2+3−4−5+6+7+8−9−10+…−2010 1+2+3-4-5+6+7+8-9-10+\ldots-2010
where each three consecutive signs ++ are followed by two signs −-.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We can write the sum as follows

∑k=0401[5k+1+5k+2+5k+3−(5k+4)−(5k+5)]=∑k=0401(5k−3)=5∑k=1401k−3⋅402=5⋅401⋅4022−3⋅402=402⋅(5⋅4012−3)=201⋅1999=401799\begin{gathered} \sum_{k=0}^{401}[5k+1+5k+2+5k+3-(5k+4)-(5k+5)] \\ =\sum_{k=0}^{401}(5k-3)=5\sum_{k=1}^{401}k-3\cdot 402 \\ =5 \cdot \frac{401 \cdot 402}{2}-3 \cdot 402 \\ =402 \cdot\left(5 \cdot \frac{401}{2}-3\right)=201 \cdot 1999=401799 \end{gathered}

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.