A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If $f(1)=2$, what is $f(8)$? A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If f(1)=2, what is f(8)?
I would like to try to tackle this problem but I need somewhere to start as I really have no idea at all on how to start.
At first I assumed that perhaps $f$ was a polynomial function, like a quadratic, but the $3xy$ term in the equation says otherwise.
Update -Thanks for everyone's help
This is what everyone has suggested. Instead of trying to find out the function, just use the f(1) value to find f(8):
$f(1+1)= f(2) = f(1) + f(1) +3(1)(1) = 7$
$f(4) = f(2) + f(2) + 3(2)(2) = 26$
$f(8) = f(4) + f(4) + 3(4)(4) = 100$
 A: Assuming you mean $f(x+y)=f(x)+f(y)+3xy$.
Hint.  Taking $x=y=1$, we have
$$f(2)=f(1)+f(1)+3=7$$
and so on.
A: Let me list out a table, assuming you mean $f(x+y)=f(x)+f(y)+3xy$:
\begin{array}{|c|c|}
x & f(x)\\
1& 2\\
2 & 7\\
4 & 26\\
8 & \boxed{100}
\end{array}
A: Setting $y=x$, we get
$$f(2x) = 2f(x) + 3x^2$$
Hence, if $x=2^n$, we get
$$f(2^{n+1}) = 2f(2^n) + 3 \cdot 2^{2n} = 2(2f(2^{n-1}) + 3 \cdot 2^{2n-2}) + 3 \cdot 2^{2n} = 4 f(2^{n-1}) + 3(2^{2n-1} + 2^{2n})$$
In general, we have
$$f(2^{n+1}) = 2^{k+1} f(2^{n-k}) + 3 \left(2^{2n} + 2^{2n-1} + \cdots +2^{2n-k}\right) = 2^{k+1}f(2^{n-k}) + 3 \cdot 2^{2n-k} \left(2^{k+1}-1\right)$$
Setting $k=n$, we get
$$f(2^{n+1}) = 2^{n+1} f(1) + 3 \cdot 2^n \cdot \left(2^{n+1}-1\right)$$
Setting $n=2$, gives you the answer.

A more direct way is to recognize that
$$f(x+1) = f(x) + f(1) + 3x$$
Similarly, we get
\begin{align}
f(x) & = f(x-1) + f(1) + 3(x-1)\\
f(x-1) & = f(x-2) + f(1) + 3(x-2)\\
f(x-2) & = f(x-3) + f(1) + 3(x-3)\\
\vdots & = \vdots\\
f(2) & = f(1) + f(1) + 3 \cdot 1
\end{align}
Using telescopic summation we get
$$f(x+1) = f(1) + x f(1) + 3 \sum_{k=1}^x k$$
Hence,
$$f(x+1) = (x+1)f(1) + \dfrac{3x(x+1)}2$$
Since $f(1)$ is given as $2$, we get
$$f(x+1) = 2(x+1) + \dfrac{3x(x+1)}2 = \dfrac{(x+1)(3x+4)}2$$
Hence,
$$f(x) = \dfrac{x(3x+1)}2$$
A: We can also set $y=1$ and solve the recursion
$$
f(x)=f(x-1)+3x-1
$$
to get that for $x\in\mathbb{Z}$,
$$
\begin{align}
f(x)
&=\sum_{k=1}^x3k-1\\
&=x(3x+1)/2
\end{align}
$$
Checking this formula against the given relation, we see that $f$ is well-defined (it doesn't depend on how we get to a particular argument):
$$
(x+y)(3(x+y)+1)/2=x(3x+1)/2+y(3y+1)/2+3xy
$$

Calculate $f(8)$:
$$
f(8)=8(3\cdot8+1)/2=100
$$
