Prove the following statement by mathematical induction.For every integer n ≥ 0, 7n − 2n is divisible by 5.Proof (by mathematical induction): Let P(n) be the following sentence.7n − 2n is divisible by 5.We will show that P(n) is true for every integer n ≥ 0.Show that P(0) is true: Select P(0) from the choices below.5 | (70 − 20)70 − 20 < 5 (70 − 20) | 55 is a multiple of 70 − 20The truth of the selected statement follows from the definition of divisibility and the fact that 70 − = 0.Show that for each integer k ≥ 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below.5 is a multiple of 7k − 2k5 is divisible by 7k − 2k 7k − 2k is divisible by 57k − 2k < 5[This is P(k), the inductive hypothesis.]We must show that P(k + 1) is true. Select P(k + 1) from the choices below.5 is a multiple of 7k + 1 − 2k + 17k + 1 − 2k + 1 < 5 7k + 1 − 2k + 1 is divisible by 55 is divisible by 7k + 1 − 2k + 1
Question
Prove the following statement by mathematical induction.For every integer n ≥ 0, 7n − 2n is divisible by 5.Proof (by mathematical induction): Let P(n) be the following sentence.7n − 2n is divisible by 5.We will show that P(n) is true for every integer n ≥ 0.Show that P(0) is true: Select P(0) from the choices below.5 | (70 − 20)70 − 20 < 5 (70 − 20) | 55 is a multiple of 70 − 20The truth of the selected statement follows from the definition of divisibility and the fact that 70 − = 0.Show that for each integer k ≥ 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below.5 is a multiple of 7k − 2k5 is divisible by 7k − 2k 7k − 2k is divisible by 57k − 2k < 5[This is P(k), the inductive hypothesis.]We must show that P(k + 1) is true. Select P(k + 1) from the choices below.5 is a multiple of 7k + 1 − 2k + 17k + 1 − 2k + 1 < 5 7k + 1 − 2k + 1 is divisible by 55 is divisible by 7k + 1 − 2k + 1
Solution
The text you provided seems to be a bit jumbled, but I understand that you're asking for a proof by mathematical induction that for every integer n ≥ 0, 7^n - 2^n is divisible by 5. Here's how you can do it:
Base Case (n=0): P(0): 7^0 - 2^0 = 1 - 1 = 0, which is divisible by 5.
Inductive Step: Assume P(k) is true, i.e., 7^k - 2^k is divisible by 5. We need to prove that P(k+1) is also true, i.e., 7^(k+1) - 2^(k+1) is divisible by 5.
7^(k+1) - 2^(k+1) = 77^k - 22^k = 57^k + 27^k - 22^k = 57^k + 2*(7^k - 2^k)
From the inductive hypothesis, we know that 7^k - 2^k is divisible by 5. Therefore, 2*(7^k - 2^k) is also divisible by 5. And obviously, 5*7^k is divisible by 5.
So, the sum of two numbers that are divisible by 5 is also divisible by 5. Hence, 7^(k+1) - 2^(k+1) is divisible by 5.
Therefore, by mathematical induction, for every integer n ≥ 0, 7^n - 2^n is divisible by 5.
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