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 Σ_{n=1}^{10} f(n) is equal to.

Leta,b,c ϵRf(1)=a+b+c=3f(2)=f(1+1)=f(1)+f(1)+1.∵f(x+y)=f(x)+f(y)+xy⟹f(2)=2f(1)+1f(3)=f(2+1)=f(2)+f(1)+2=2f(1)+1+f(1)+2⟹f(3)=3f(1)+3f(4)=f(3+1)=f(3)+f(1)+3=3f(1)+3+f(1)+3⟹f(4)=4f(1)+610∑n=1f(n)=f(1)+f(2)+f(3)+−−−−−−−−+f(10)=f(1)+2f(1)+1+3f(1)+3+4f(1)+6+5f(1)+10−−−−−−−−=f(1)[1+2+3+4+−−−−−+10]+(1+3+6+10+15+21+28+36+45)=f(1)(55)+(165)=3×55+165=330∴10∑n=1f(n)=330Hence, option A is correct.