Question

Given ax2 + bx+c > 0; bx2 + cx + a > Oand cx2 + ax + b > Ofor all
a² +6² +2
X E Rwhere a + b + cand a, b, c E R. Now can not take values
ab + bctca
(A)
7
9
(B)
7
(C)
16
3
(D)
2​

Answer 1

Answer:

n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)

⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)