11.3. The function $f(x)$ is such that for all values of $x$ the equality $f(x+1)=f(x)+$ $+2x+3$ holds. It is known that $f(0)=1$. Find $f(2012)$.
Answer: $4052169=2013^{2}$.

Write the condition of the problem as: $f(x+1)-f(x)=2 x+3$. Substitute for $x$ the numbers $0,1,2$, ..., 2011. We get: $f(1)-f(0)=2 \cdot 0+3, f(2)-f(1)=2 \cdot 1+3, \ldots, f(2012)-f(2011)=2 \cdot 2011+3$.

Add the obtained equations term by term: $f(2012)-f(0)=2 \cdot(1+2+\ldots+2011)+3 \cdot 2012$. Considering that $1+2+\ldots+2011=\frac{2012 \cdot 2011}{2}$, we get that $f(2012)=1+2011 \cdot 2012+3 \cdot 2012=$ $=1+2 \cdot 2012+2012^{2}=2013^{2}=4052169$.

A schoolchild has the right to write the answer in either of the two specified forms.

+ complete justified solution

$\mp$ by considering individual examples, it is noticed but not proven that the desired value is a square, and the correct answer is given

- only the answer is given