Question

1+2+3+4.............+414+415=

Answer 1

$1+2+3+4+...........+414+415= \\ \frac{415*(415+1)}{2}= \\ \frac{415*416}{2}= \\ \frac{172640}{2}=86320$
formula de calcul:
$\frac{n*(n+1)}{2}$
unde n este ultimul termen al sirului

Answer 2

$\displaystyle 1+2+3+4+...+414+415= \frac{415(415+1)}{2} = \frac{415 \cdot 416}{2} = \\ \\ = \frac{172640}{2} =86320$