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

Question

Let a, b, c ∈ R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f (x + y) = f (x) + f (y) + xy, ∀ x, y ∈ R, then displaystyle∑10n = 1 f(n) is equal to : (A) 165 (B) 190 (C) 255 (D) 330

Tardigrade Admin · Accepted Answer

As, f (x + y) = f (x) + f (y) + xy Given, f (1) = 3 Putting, x = y = 1 ⇒ f (2) = 2f (1) +1 = 7 Similarly, x = 1, y = 2 ⇒ f (3) = f (1) + f (2) + 2 = 12 Now, displaystyle ∑n=110f(n) = f (1) + f (2) + f (3) + ... + f (10) = 3 + 7 +12 + 18 + ... = S (let) Now, Sn = 3 + 7 + 12 + 18 + ... + tn Again, Sn = 3 + 7 + 12 + ... + tn-1 + tn We ge, tn = 3 + 4 + 5+ ... n terms = (n(n+5)/2) i.e., Sn = displaystyle ∑n=1ntn = (1/2) ∑ n2+5∑ n = (n(n+1)(n+8)/6) So, S10 = (10×11×18/6) = 330

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

165

190

255

330

Q. Let $a, b, c \in R$ . If $f (x) = a x^{2} + b x + c$ is such that $a + b + c = 3$ and $f (x + y) = f (x) + f (y) + x y, \forall x, y \in R,$ then $n = 1 \sum 10 f (n)$ is equal to :