Question

(6+12 +18+...+300) la puterea 3/(3 +6 +9 +....+150)

Answer 1

[tex]\dfrac{(6+12+_{\dots}+300)^3}{3+6+9+_{\dots}+150}=\dfrac{6^3\cdot (1+2+3+_{\dots}+50)^3}{3(1+2+3+_{\dots}+50)}=\\ \dfrac{216}{3}\cdot (1+2+3+_{\dots}+50)^2=72\cdot \left(\dfrac{50\cdot 51}{2}\right)^2=72\cdot (25\cdot 51)^2=\\ =72\cdot 625\cdot 2601=\boxed{117045000}[/tex]

Answer 2

[tex]\dfrac{(6+12 +18+...+300)^3}{3+6+9+...+150} = \dfrac{\Big[2\cdot (3+6+9+...+150)\Big]^3}{3+6+9+...+150} = \\ \\ \\ = \dfrac{2^3\cdot (3+6+9+...+150)^3}{3+6+9+...+150} = 2^3\cdot (3+6+9+...+150)^2 = \\ \\ =2^3\cdot \Big[3\cdot ( 1+2+3+...+50)\Big]^2 = 2\cdot 2^2\cdot \left(3\cdot \dfrac{50\cdot (50+1)}{2}\right)^2 = \\ \\ = 2\cdot \left(2\cdot 3\cdot \dfrac{50\cdot 51}{2}\right)^2 = 2\cdot (3\cdot 50\cdot 51)^2 = 2\cdot 7650^2 = 2\cdot 58522500 = \\ \\ = 117045000[/tex]