Originally Posted by

**Moo** Hello,

Sum of all even numbers between 2 and 200 :

$\displaystyle 2+4+ \dots + 200=2(1+2+\dots+100)=2 \sum_{n=1}^{100} n$

Among these (integers from 1 to 100), remove the ones in the form $\displaystyle 3k$

Sum of all multiples of 3 between 1 and 100 :

$\displaystyle 3+6+\dots+99=3(1+2+\dots+33)=3 \sum_{n=1}^{33} n$

So $\displaystyle S=2 \left( \sum_{n=1}^{100} n-3 \sum_{n=1}^{33} n\right)$

I misunderstood the question, thanks for helping me out.

$\displaystyle S=2 \left(\frac{100 \times 101}{2}-3 \cdot \frac{33 \times 34}{2} \right)$

$\displaystyle S=2 \left(50 \times 101-3 \times 33 \times 17 \right)$

$\displaystyle S=6734$