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

We are told that $$f(x)=ax^{2}+bx+c$$ with real constants $$a,b,c$$ satisfies two conditions:

1. $$a+b+c=3$$.

2. $$f(x+y)=f(x)+f(y)+xy$$ for every real pair $$x,y$$.

First we expand the left-hand side of the functional equation. Taking $$f(x)=ax^{2}+bx+c$$ we get

$$ \begin{aligned} f(x+y)&=a(x+y)^{2}+b(x+y)+c \\ &=a(x^{2}+2xy+y^{2})+bx+by+c \\ &=ax^{2}+2axy+ay^{2}+bx+by+c. \end{aligned} $$

Next we write the right-hand side, namely $$f(x)+f(y)+xy$$. Computing separately,

$$ \begin{aligned} f(x)+f(y) &=\bigl(ax^{2}+bx+c\bigr)+\bigl(ay^{2}+by+c\bigr) \\ &=ax^{2}+ay^{2}+bx+by+2c. \end{aligned} $$

Adding the extra $$xy$$ term required by the equation, we obtain

$$f(x)+f(y)+xy=ax^{2}+ay^{2}+bx+by+2c+xy.$$

Because the functional equation holds for all real $$x,y$$, the two expanded expressions must match term by term. We therefore equate coefficients of each power of $$x$$ and $$y$$:

• Coefficient of $$x^{2}$$: $$a=a$$ (already consistent).

• Coefficient of $$y^{2}$$: $$a=a$$ (already consistent).

• Coefficient of the mixed term $$xy$$: we have $$2a$$ on the left and $$1$$ on the right, so

$$2a=1 \;\;\Longrightarrow\;\; a=\dfrac12.$$

• Constant term: left side has $$c$$ whereas right side has $$2c$$, hence

$$c=2c \;\;\Longrightarrow\;\; c=0.$$

With $$a=\dfrac12$$ and $$c=0$$ found, we use the earlier restriction $$a+b+c=3$$ to determine $$b$$:

$$ \frac12 + b + 0 = 3 \;\;\Longrightarrow\;\; b = 3-\frac12 = \frac52. $$

Thus the concrete quadratic is

$$f(x)=\frac12x^{2}+\frac52x.$$

Now we need the sum $$\displaystyle\sum_{n=1}^{10}f(n)$$. Substituting our explicit form,

$$ \sum_{n=1}^{10}f(n) =\sum_{n=1}^{10}\left(\frac12n^{2}+\frac52n\right) =\frac12\sum_{n=1}^{10}n^{2}+\frac52\sum_{n=1}^{10}n. $$

We recall the standard summation formulas (stated here for clarity):

• $$\displaystyle\sum_{n=1}^{N} n = \frac{N(N+1)}{2}.$$

• $$\displaystyle\sum_{n=1}^{N} n^{2} = \frac{N(N+1)(2N+1)}{6}.$$

Taking $$N=10$$, we compute each:

$$ \sum_{n=1}^{10} n = \frac{10\cdot11}{2}=55, \qquad \sum_{n=1}^{10} n^{2} = \frac{10\cdot11\cdot21}{6}=385. $$

Substituting these values back, we get

$$ \begin{aligned} \sum_{n=1}^{10}f(n) &=\frac12\,(385)+\frac52\,(55) \\ &=\frac{385}{2}+\frac{275}{2} \\ &=\frac{660}{2}=330. \end{aligned} $$

Hence, the correct answer is Option A.