# Task 3. (12 points)

A numerical sequence is such that $x_{n}=\frac{x_{n-1}}{5}+4$ for all $n \geq 2$. Find $x_{2017}$ if $x_{1}=6$.

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# Solution.

$x_{2}=\frac{x_{1}}{5}+4=\frac{6}{5}+4=5+\frac{1}{5}$.

$x_{3}=\frac{x_{3}}{5}+4=\frac{5+\frac{1}{5}}{5}+4=5+\frac{1}{5^{2}}$.

We will prove by induction that $x_{n}=5+\frac{1}{5^{1-n}}$ for all natural $n$.

1) $n=1: \quad x_{1}=5+\frac{1}{5^{1-1}}=6$ - true.
2) Suppose the statement is true for $n=k: x_{k}=5+\frac{1}{5^{1-k}}$.
3) $n=k+1: x_{k+1}=\frac{x_{k}}{5}+4=\frac{5+\frac{1}{5^{1-k}}}{5}+4=5+\frac{1}{5^{1-(k+1)}}$.

Since $x_{n}=5+\frac{1}{5^{1-n}}$, then $x_{2017}=5+\frac{1}{5^{-2016}}$.

Answer: $5+\frac{1}{5^{-2016}}$.

| Full solution. | + | 12 |
| :---: | :---: | :---: |
| The first few terms of the sequence are written out. Additional justifications are not provided or the provided justifications are incorrect. The answer is correct. | $+/ 2$ | 6 |
| The first few terms of the sequence are written out as the sum of 5 and a fraction. The answer is missing or incorrect. | $\mp$ | 2 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score |  | 12 |