Question

Calculați sumele:
a) 1+2+3+...+99
b) 2+4+6+...+198
Urgent!!

Answer 1

$\displaystyle a).1+2+3+...+99= \frac{99(99+1)}{2} = \frac{99 \times 100}{2} = \frac{9900}{2} = 4950 \\ \\ b).2+4+6+...+198=2(1+2+3+...+99)=2 \times \frac{99(99+1)}{2} = \\ \\ =2 \times \frac{99 \times 100}{2} =\not2 \times \frac{9900}{\not2} =9900$