momo1729 wrote: Compute the sum $S=1+2+3-4-5+6+7+8-9-10+...-2010$ where every three consecutive $+$ are followed by two $-$. $S=\sum_{n=1}^{402}((5n-4)+(5n-3)+(5n-2)-(5n-1)-(5n))$ $=\sum_{n=1}^{402}(5n-8)$ $S=5\left(\sum_{n=1}^{402}n\right)-8\times 402$ $S=5\frac{402\times 403}2-3216$ $\boxed{S=401799}$ Certainly not an olympiad exercise

Well, a very easy one... It's from a Moroccan Olympiad and is mainly for elimination purposes.

See here for a related problem.

nice one but non olympiad problem))

There is another way to solve this problem that I don't think was already up, and it doesn't involve sigma: I would probably break this into two sequences: 1 + 2 + 3 + 6 + 7 + 8 + 11 + 12 + 13 + ....... + 2006 + 2007 + 2008 and -4 - 5 - 9 - 10 - 14 - 15 - ........ - 2009 - 2010 I would then pair up the sets of positive and negative numbers. For the first set, {-5,-4,1,2,3}, I wish to leave off the one for now: -5 + 3 - 4 + 2, which equals -4. Note that for the last four numbers in each of these sets, the sum is -4, and then the extra number (1,6,11,16,etc.) is added on. So, since there are of these 402 sets of numbers, the sequence could be rewritten as: (-4 x 402) + (1 + 6 + 11 + 16 + 21 + ........ + 2006) With some simple and obvious multiplication, this is broken up to -1608 + (1 + 6 + 11 + 16 + 21 + ......... + 2006). Then, the new sequence can be simplified into a positive number: There are 201 of these pairs (and 402 numbers in the new sequence). The last and first number add up to 2007, the second-to-last and the second to 2007, and so on... So, it is now -1608 + (2007)(201) To solve without a calculator: -1608 + (2007)(200) + 201 -1407 + 401400, which equals 399993. Is this right? I think others were getting a different answer?

I would not judge the choice of problems for the Olympiad as it could be their (or one of the first)first year(s) and therefore they are not very used to harder problems. Hopefully though, they will get harder ones in the next competitions.

It's a relatively easy problem that I like. Nice work creator of problem!

Let $U_{n}=n+(n+1)+(n+2)-(n+3)-(n+4)=n-4 $ $S=U_{1}+U_{6}+..+U_{2006}$ $S=\sum \limits_{k=0}^{401} U_{5k+1}$ $S=\sum \limits_{k=0}^{401} 5k-3$ $S=5 \sum \limits_{k=0}^{401} k -3*402$ $S= 5* \frac{401*402}{2} -3*402$ $$S=401799$$

Kingofmath101 wrote: So, it is now -1608 + (2007)(201) To solve without a calculator: -1608 + (2007)(200) + 201 -1407 + 401400, which equals 399993. Is this right? I think others were getting a different answer? You made a mistake here. $(2007)(201)$ is $(2007)(200)+2007$.