Step 1: Find the sum and product of the roots. 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: $\alpha \beta = 36/1 = 36$ Step 2: Define a recurrence relation for $S_n = \alpha^n + \beta^n$. 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: $$S_n = 14S_{n-1} - 36S_{n-2} \quad \text{for } n \ge 2$$ Step 3: Prove by induction that $S_n$ is divisible by $2^n$. 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: $$S_{k+1} = 14S_k - 36S_{k-1}$$ Substitute the inductive hypothesis: $$S_{k+1} = 14(c_k \cdot 2^k) - 36(c_{k-1} \cdot 2^{k-1})$$ $$S_{k+1} = (2 \cdot 7)(c_k \cdot 2^k) - (4 \cdot 9)(c_{k-1} \cdot 2^{k-1})$$ $$S_{k+1} = 7 \cdot c_k \cdot 2^{k+1} - 9 \cdot c_{k-1} \cdot 2^2 \cdot 2^{k-1}$$ $$S_{k+1} = 7 \cdot c_k \cdot 2^{k+1} - 9 \cdot c_{k-1} \cdot 2^{k+1}$$ Factor out $2^{k+1}$: $$S_{k+1} = 2^{k+1}(7c_k - 9c_{k-1})$$ 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 $\boxed{\text{Shown that } \alpha^n + \beta^n \text{ is divisible by } 2^n \text{ for } n=1,2,3,...}$