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

From the first equation 
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+xy$$
$$2axy=c+xy$$
Also, the summation will be
$$a(1^2+2^2+3^2...10^2)+b(1+2+3+4...+10)+10c$$
$$=385a+55b+10c$$
$$375a+45b+30$$ 
I don’t know what to do with the x and y terms. How do  I proceed?

You have that
$$f(x+y)=f(x)+f(y)+xy \tag{1}\label{eq1A}$$
for all $x,y\in R$. Your simplification of \eqref{eq1A} gives
$$2axy=c+xy \implies (2a-1)xy = c \tag{2}\label{eq2A}$$
Since this is true for every real $x$ and $y$, this means that $2a - 1 = 0 \implies a = \frac{1}{2}$ and $c = 0$. This thus gives that $b = 3 - a - c = \frac{5}{2}$. Can you finish the rest?
