1. Give an example of a quadratic trinomial $P(x)$ such that for any $x$, the following equation holds:
$$
P(x)+P(x+1)+\cdots+P(x+10)=x^{2} .
$$
1. $\frac{x^{2}-10 x+15}{11}$.

Let the quadratic trinomial be
$$
P(x)=a x^{2}+b x+c \text {. }
$$

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

According to the problem, the above expression should equal $x^{2}$.
By comparing the coefficients of corresponding terms, we get the system of equations
$$
\left\{\begin{array}{l}
11 a=1, \\
110 a+11 b=0, \\
385 a+55 b+11 c=0 .
\end{array}\right.
$$

Solving these, we get $a=\frac{1}{11}, b=-\frac{10}{11}, c=\frac{15}{11}$.