Discrete Math
posted by Francesca .
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!

Discrete Math 
MathMate
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?
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