Evaluate the sum
where each three consecutive signs are followed by two signs .
Solution
We can write the sum as follows
\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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