Question

1+2+...+64=
1+2+...+215=
1+2+...+625=
1+2...+750=
Va rog ajutatima

Answer 1

$\displaystyle a).1+2+...+64= \frac{64(64+1)}{2} = \frac{64 \times 65}{2} = \frac{4160}{2} =2080 \\ \\ b).1+2+...+215= \frac{215(215+1)}{2} = \frac{215 \times 216}{2} = \frac{46440}{2} =23220 \\ \\ c).1+2+...+625= \frac{625(625+1)}{2} = \frac{625 \times 626}{2} = \frac{391250}{2} =195625 \\ \\ d).1+2+...+750= \frac{750(750+1)}{2} = \frac{750 \times 751}{2} = \frac{563250}{2} =281625$