Question

1+2+3+.......+147
1+2+3+4+5+.........+201
1+2+3+...............+1002

Answer 1

$\displaystyle 1+2+3+...+147= \frac{147(147+1)}{2} = \frac{147 \times 148}{2} = \frac{21756}{2} =10878 \\ \\ 1+2+3+4+5+...+201= \frac{201(201+1)}{2} = \frac{201 \times 202}{2} = \\ \\ = \frac{40602}{2} =20301 \\ \\ 1+2+3+...+1002= \frac{1002(1002+1)}{2} = \frac{1002 \times 1003}{2} = \\ \\ = \frac{1005006}{2} =502503$