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

Step 1: Find the sum and product of the roots. Given the quadratic equation x^2 - 14x + 36 = 0. Let and be the roots. Using Vieta's formulas: Sum of roots: + = -(-14)/1 = 14 Product of roots: = 36/1 = 36 Step 2: Define a recurrence relation for S_n = ^n + ^n. Since and are roots of x^2 - 14x + 36 = 0, they satisfy the equation: ^2 - 14 + 36 = 0 ^2 = 14 - 36 ^2 - 14 + 36 = 0 ^2 = 14 - 36 Multiply the first equation by ^n-2 and the second by ^n-2 (for n 2): ^n = 14^n-1 - 36^n-2 ^n = 14^n-1 - 36^n-2 Adding these two equations: ^n + ^n = 14(^n-1 + ^n-1) - 36(^n-2 + ^n-2) Let S_n = ^n + ^n. The recurrence relation is: S_n = 14S_n-1 - 36S_n-2 for n 2 Step 3: Prove by induction that S_n is divisible by 2^n. Base Cases: For n=1: S_1 = + = 14. 2^1 = 2. 14 is divisible by 2 (14 = 2 × 7). So, S_1 is divisible by 2^1. For n=2: S_2 = ^2 + ^2 = (+)^2 - 2 = (14)^2 - 2(36) = 196 - 72 = 124. 2^2 = 4. 124 is divisible by 4 (124 = 4 × 31). So, S_2 is divisible by 2^2. Inductive Hypothesis: Assume that for some integer k 1, S_k is divisible by 2^k and S_k-1 is divisible by 2^k-1. This means S_k = c_k · 2^k and S_k-1 = c_k-1 · 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 · 2^k) - 36(c_k-1 · 2^k-1) S_k+1 = (2 · 7)(c_k · 2^k) - (4 · 9)(c_k-1 · 2^k-1) S_k+1 = 7 · c_k · 2^k+1 - 9 · c_k-1 · 2^2 · 2^k-1 S_k+1 = 7 · c_k · 2^k+1 - 9 · c_k-1 · 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, ^n + ^n is divisible by 2^n for all n=1,2,3,.... The final answer is Shown that ^n + ^n is divisible by 2^n for n=1,2,3,...