Provide an example of a quadratic trinomial $P(x)$ such that for any $x$ the equality $P(x)+P(x+1)+\cdots+P(x+10)=x^{2}$ holds.
Let the desired polynomial be $P(x)=a x^{2}+b x+c$. Then

$$
\begin{aligned}
P(x)+ & P(x+1)+\cdots+P(x+10)= \\
= & a\left(x^{2}+(x+1)^{2}+\cdots+(x+10)^{2}\right)+ \\
& \quad+b(x+(x+1)+\cdots+(x+10))+11 c= \\
= & a\left(11 x^{2}+(2+4+\cdots+20) x+\left(1+2^{2}+\cdots+10^{2}\right)\right)+ \\
+ & b(11 x+1+2+\cdots+10)+11 c= \\
= & 11 a x^{2}+110 a x+385 a+11 b x+55 b+11 c= \\
= & 11 a x^{2}+(110 a+11 b) x+(385 a+55 b+11 c) .
\end{aligned}
$$

We obtain the equality of quadratic trinomials

$$
11 a x^{2}+(110 a+11 b) x+(385 a+55 b+11 c) \quad \text { and } \quad x^{2}
$$

This is equivalent to the equality of coefficients, that is, the system of equations

$$
\left\{\begin{array}{l}
11 a=1 \\
110 a+11 b=0 \\
385 a+55 b+11 c=0
\end{array}\right.
$$

which has a unique solution $a=\frac{1}{11}, b=-\frac{10}{11}, c=\frac{15}{11}$.

## Answer

$\frac{x^{2}-10 x+15}{11}$