Problem 2. 2-1. Find the minimum value of the function

$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-100)^{2}
$$

If the result is a non-integer, round it to the nearest integer and write it as the answer.
Solution. Knowledge of arithmetic progression is required. It turns out to be a quadratic function

$f(x)=51 x^{2}-2(2+4+6+\ldots+100) x+(2^{2}+4^{2}+6^{2}+\ldots+100^{2})=51 x^{2}-2 \cdot 50 \cdot 51 x+4 \cdot(1^{2}+2^{2}+3^{2}+\ldots+50^{2})$.

The minimum is achieved at the point $x_{0}=50$. In this case,

$f(50)=50^{2}+48^{2}+46^{2}+\ldots+2^{2}+0^{2}+2^{2}+4^{2}+\ldots+50^{2}=2 \cdot(2^{2}+4^{2}+\ldots+50^{2})=8 \cdot(1^{2}+2^{2}+3^{2}+\ldots+25^{2})$.

Now we need the formula

$$
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$

(it is known, but in principle, it can be derived). Then $f(50)=8 \cdot \frac{25 \cdot 26 \cdot 51}{6}=17 \cdot 26 \cdot 100=44200$.

Answer: $44'200$.