Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this sequence can contain.
# Solution.

Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that an arithmetic progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equation $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.

Indeed,

$a_{k}=2 k, \quad a_{k+1}=2 k+2, a_{k+2}=2 k+4$, or by the formula

$a_{k+2}=3 a_{k+1}-2 a_{k}-2=3(2 k+2)-2 \cdot 2 k-2=2 k+4$

The sequence containing 2022: 4, 6,

This finite sequence can contain all even three-digit numbers from 100 to 999. Among them, the numbers divisible by 4 are: 100, 104, 108, 112, ..., 992, 925, 950, 996 - 25 in each of the nine hundreds, i.e., 225 numbers.

Answer. 225.