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?

Hint:
$a+b+c=3 \equiv f(1)=3$. Using recurrence relation, $f(2)=2f(1)+1=7$, $f(3)=12$, $f(4)=18$. Notice the pattern. 
$$\begin{aligned}a+b+c&=3\\4a+2b+c&=7\\9a+3b+c&=12\end{aligned}$$
This gives $a=1/2$, $b=5/2$, $c=0$. Can you proceed? 
