Use mathematical induction to prove the truth of each of the following assertions for all n ≥1.

5^2n – 2^5n is divisible by 7

If n = 1, then 5^2(1) - 2^5(1) = -7, which is divisible by 7. For the inductive case, assume k ≥ 1, and the result is true for n = k; that is 7 | (5^2k + 2^5k). Use the assumption to prove n = k + 1, in other words, 5^(2(k + 1)) - 2^(5(k + 1)) is divisible by 7. Now,

5^(2(k + 1)) - 2^(5(k + 1))
= 5^(2k + 2) - 2(5k + 5)
= 5^(2k) · 5^2 - 2^(5k) · 2^5
= 25 · 5^(2k) - 32 · 2^(5k)
= IDK what to do from here. . .

Any suggestions? Thank you again!

To continue the proof using mathematical induction, we need to show that 25 · 5^(2k) - 32 · 2^(5k) is divisible by 7.

Starting with the expression:

25 · 5^(2k) - 32 · 2^(5k)

We can rewrite 25 as 7*3 + 4, and 32 as 7*4 + 4:

(7*3 + 4) · 5^(2k) - (7*4 + 4) · 2^(5k)

Expanding this expression:

= 7*3·5^(2k) + 4 · 5^(2k) - 7*4 · 2^(5k) - 4 · 2^(5k)

Factoring out 7 from the first and third terms, and 4 from second and fourth terms:

= 7(3·5^(2k) - 4·2^(5k)) + 4(5^(2k) - 2^(5k))

Now, we want to show that this expression is divisible by 7.

By the assumption, we know that 7 | (5^2k + 2^5k), which means that 5^(2k) - 2^(5k) is divisible by 7. In other words, there exists an integer m such that:

5^(2k) - 2^(5k) = 7m

Substituting this into our expression:

= 7(3·5^(2k) - 4·2^(5k)) + 4(7m)

= 7(3·5^(2k) - 4·2^(5k) + 4m)

Now, we have shown that the expression is a multiple of 7, which means it is divisible by 7.

Therefore, we have proven that 5^(2n) - 2^(5n) is divisible by 7 for all n ≥ 1 using mathematical induction.

Let's continue:

25 · 5^(2k) - 32 · 2^(5k)
=25*(5^(2k)-2^(5k) -7*2^(5k)

Now ask yourself:
A. Is (5^(2k)-2^(5k) divisible by 7, and why?
B. Is -7*2^(5k) divisible by 7, and why?