Question

calculati:
a) (1+2+...+100):(-101)
b) (-2-4-6-...-1000):(-2)
c) (-3-6-9-...-3000):(1+2+3+....+1000)
VA ROG UN RASPUNS CAT MAI RAPID SI COMPLEX!!!!
DAU COROANA​

Answer 1

Hey! Pls coroana :)

trebuie sa stii formula :1+2+3+.....+n=n(n+1)/2'deci

1+2+3+...+100=100x101/2, care impartit la (-101)=-50

b) dai factor comun pe -2, rezulta:  -2(1+2+3+....+500):(-2)=

-2x500x501/2:(-2)=250x501

c) -3(1+2+3+...+1000):1000x1001/2=-3x1000x1001/2:500x1001=

-3

Succes !

Answer 2

 

$\displaystyle\bf\\a)\\(1+2+...+100):(-101)=\\\\=\frac{100(100+1)}{2}:(-101)=\\\\=\frac{100(100+1)}{2\times(-101)}=\\\\=-\frac{100\times101}{2\times101}=-\frac{100}{2}=\boxed{\bf-50}\\\\\\b)\\(-2-4-6-...-1000):(-2)=\\\\=\frac{(-2-4-6-...-1000)}{-2}=\\\\=\frac{-2(1+2+3+...+500)}{-2}=\\\\=1+2+3+...+500=\frac{500(500+1)}{2}=250\times501=\boxed{\bf125250}$

.

$\displaystyle\bf\\c)\\(-3-6-9-...-3000):(1+2+3+....+1000)=\\\\=\frac{(-3-6-9-...-3000)}{(1+2+3+....+1000)}=\\\\\\=\frac{-3(1+2+3+...+1000)}{(1+2+3+....+1000)}=\boxed{\bf-3}$