Given the quadratic equation x² − 14x + 36 = 0.

Given the quadratic equation $x^2 - 14x + 36 = 0$.
Let $\alpha$ and $\beta$ be the roots.
Using Vieta's formulas:
Sum of roots: $\alpha + \beta = -(-14)/1 = 14$
Product of roots: $\alpha \beta = 36/1 = 36$

Since $\alpha$ and $\beta$ are roots of $x^2 - 14x + 36 = 0$, they satisfy the equation:
$\alpha^2 - 14\alpha + 36 = 0 \implies \alpha^2 = 14\alpha - 36$
$\beta^2 - 14\beta + 36 = 0 \implies \beta^2 = 14\beta - 36$
Multiply the first equation by $\alpha^{n-2}$ and the second by $\beta^{n-2}$ (for $n \ge 2$):
$\alpha^n = 14\alpha^{n-1} - 36\alpha^{n-2}$
$\beta^n = 14\beta^{n-1} - 36\beta^{n-2}$
Adding these two equations:
$\alpha^n + \beta^n = 14(\alpha^{n-1} + \beta^{n-1}) - 36(\alpha^{n-2} + \beta^{n-2})$
Let $S_n = \alpha^n + \beta^n$. The recurrence relation is:

Base Cases:
For $n=1$:
$S_1 = \alpha + \beta = 14$.
$2^1 = 2$.
14 is divisible by 2 ($14 = 2 \times 7$). So, $S_1$ is divisible by $2^1$.

For $n=2$:
$S_2 = \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (14)^2 - 2(36) = 196 - 72 = 124$.
$2^2 = 4$.
124 is divisible by 4 ($124 = 4 \times 31$). So, $S_2$ is divisible by $2^2$.

Inductive Hypothesis:
Assume that for some integer $k \ge 1$, $S_k$ is divisible by $2^k$ and $S_{k-1}$ is divisible by $2^{k-1}$.
This means $S_k = c_k \cdot 2^k$ and $S_{k-1} = c_{k-1} \cdot 2^{k-1}$ for some integers $c_k$ and $c_{k-1}$.

Inductive Step:
We need to show that $S_{k+1}$ is divisible by $2^{k+1}$.
Using the recurrence relation:

Substitute the inductive hypothesis:

Factor out $2^{k+1}$:

Since $c_k$ and $c_{k-1}$ are integers, $7c_k - 9c_{k-1}$ is also an integer.
Therefore, $S_{k+1}$ is divisible by $2^{k+1}$.

By the principle of mathematical induction, $\alpha^n + \beta^n$ is divisible by $2^n$ for all $n=1,2,3,...$.

The final answer is $\boxed{\text{Shown that } \alpha^n + \beta^n \text{ is divisible by } 2^n \text{ for } n=1,2,3,...}$