$3+$ [ Arithmetic progression $]$ 

The function $f(x)$ is such that for all values of $x$ the equality $f(x+1)=f(x)+2 x+3$ holds. It is known that $f(0)=$ 1. Find $f(2012)$.
$f(x+1)-f(x)=2x+3$. Substitute $x$ with the numbers $0,1,2, \ldots, 2011: f(1)-f(0)=2 \cdot 0+3, f(2)-f(1)=2 \cdot 1+3, \ldots$, $f(2012)-f(2011)=2 \cdot 2011+3$. By adding these equations term by term, we get $f(2012)-f(0)=2 \cdot(1+2+\ldots+2011)+$ $3 \cdot 2012=2012 \cdot 2011+3 \cdot 2012$. Therefore, $f(2012)=1+2 \cdot 2012+2012^{2}=2013^{2}$.

## Answer

$4052169=2013^{2}$.
[ Decimal number system $]$
Problem