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 $a+b+c=3$, we have $$f(1)=a+b+c=3$$
and from $f(x+y)=f(x)+f(y)+xy$, we have
$$f(x+1)=f(x)+f(1)+x=f(x)+x+3$$
therefore
$$\sum_{n=1}^{10}f(n+1)=\sum_{n=1}^{10} (f(n)+n+3)$$
or $$\sum_{n=2}^{11}f(n)=\sum_{n=1}^{10}f(n)+\sum_{n=1}^{10}(n+3)$$
can you proceed?
