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?

From your third step you get $c=0$ and $a=1/2, \implies 1/2+b+0=3 \implies b=5/2$.
Next, $f(n)=n^2/2+5n/2$, then
$$S_m=\sum_{n=1}^{m} (n^2/2+5n/2)= \frac{m(m+1)(2m+1)}{12}+\frac{5m(m+1)}{4}$$
$$\implies S_m=\frac{m(m+1)(m+8)}{6}$$
then $$S_{10}=10.11.18/6=330$$
