Question

Let a,b,c∈R. If f(x) = ax² + bx + c is such that a + b + c = 3 and f x(x + y) = f(x) + f(y)y + xy, ∀ x,y ∈ R, then ∑ f(n) for n ∈[n = 1,10] is equal to:

(A) 165

(B) 190

(C) 255

(D) 330

Answer 1

Correct option  (D) 330

We have f(x) = x² + bx + c

f(1) = a + b + c = 3

Now,

f(x + y) = f(x) + f(y) + xy

Substituting y = 1, we get

f(x + 1) = f(x) + f(1) + x

f(x + 1) = f(x) + x + 3

Now, f(2) = 7 and f(3) = 12

Therefore

S_n = 3 + 7 + 12 + .... + t_n ....(1)

S_n = 3 + 7 + .... + t_{n - 1} + S_n ....(2)

On subtracting Eq. (2) from Eq. (1), we get

t_n = 3 + 4 + 5 + … upto n terms.

On subtracting Eq. (2) from Eq. (1), we get

t_n = 3 + 4 + 5 + … upto n terms.

Therefore,

On further simplification, we get Sn = 330.