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?
 A: 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?
A: 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?
A: 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? 
A: 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 $$S_{10}=10.11.18/6=330$$
