2-4. Find the minimum value of the function

$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-104)^{2}
$$

If the result is a non-integer, round it to the nearest integer and write it in the answer.
Solution. $f(x)=53 x^{2}-2(2+4+\ldots 104) x+4\left(1^{2}+2^{2}+\ldots+52^{2}\right)=53 x^{2}-2 \cdot 52 \cdot 53 x+4 \cdot \frac{52 \cdot 53 \cdot 105}{6}$ (since $\left.1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\right)$.

The minimum of the function $f$ is achieved at the point $x_{0}=52 . f(52)=53 \cdot 52^{2}-2 \cdot 53 \cdot 52^{2}+2 \cdot 52 \cdot 53 \cdot 35=$ $52 \cdot 53(70-52)=49608$.

Answer: 49608