Question

Aratati ca numarul 9*(5+10+15+...+200) : 41 = patrat perfect

Answer 1

modul 1:
9*(5+10+15+.........+200):41=
mai intai vom calcula 5+10+15+........+200
5+10+15+........+200=(200+5)*(numarul de termeni):2
numarul de termeni=(200-5):5+1=195:5+1=39+1=40
(200+5)*40:2=205*20=4100
9*4100:41=36900:41=900=30*30
modul 2:
9*(5+10+15+.........+200):41=
mai intai vom calcula 5+10+15+........+200
5+10+15+.........+200=5*(1+2+3........+40)=5*[40*(40+1):2]=5*(20*41)=5*820=4100
9*4100:41=36900:41=900=30*30

Answer 2

$9 \cdot (5+10+15+...+200):41=9 \cdot 5(1+2+3+...+40):41= \\ =45(1+2+3+...+40):41=45 \cdot \frac{40(40+1)}{2} :41= \\ =45 \cdot \frac{40 ~\cdot ~41}{2} :41=45 \cdot \frac{1640}{2} :41=45 \cdot 820:41=45 \cdot 20=900=30^2$