Question

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

Answer 1

$\frac{ {(6 + 12 + 18 + ... + 300)}^{3} }{3 + 6 + 9 + ... + 150} = \frac{{[6(1 + 2 + 3 + ... + 50)] }^{3} }{3(1 + 2 + 3 + ... + 50)}$

$= \frac{ {6}^{3} \times {(1 + 2 + 3 + ... + 50)}^{3} }{3(1 + 2 + 3 + ... + 50)} = \frac{216 {(1 + 2 + 3 + ... + 50)}^{3} }{3(1 + 2 + 3 + ... + 50)}$

$= \frac{216}{3} \times \frac{ {(1 + 2 + 3 + ... + 50)}^{3} }{1 + 2 + 3 + ... + 50)}$

$= 72 \times {(1 + 2 + 3 + ... + 50)}^{3 - 1}$

$= 72 \times {(1 + 2 + 3 + ... + 50)}^{2}$

$= 72 \times {( \frac{50(50 + 1)}{2} )}^{2}$

$= 72 \times {( \frac{50 \times 51}{2} )}^{2}$

$= 72 \times {(25 \times 51)}^{2}$

$= 72 \times {1275}^{2}$

$= 72 \times 1625625$

$= 117045000$