Question

Find the sum of the following series

3 + 6 + 9 + ... + 96

Answer 1

3 + 6 + 9 + ... + 96

= 3(1 + 2 + 3 + ... + 32)

= ${\displaystyle 3\left[\sum _1^{32}\text{n}\right]}$

= ${\displaystyle 3{\left[\frac{\text{n}(\text{n}+1)}{2}\right]}_{\text{n}=32}}$

= ${\displaystyle 3\left[\frac{32\left(33\right)}{2}\right]}$

= 1584