Sunday

April 20, 2014

April 20, 2014

Posted by **Francesca** on Wednesday, March 23, 2011 at 2:55pm.

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**, Thursday, March 24, 2011 at 9:18amLet'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?

**Related Questions**

Discrete Math - Use mathematical induction to prove the truth of each of the ...

AP Calc - Use mathematical induction to prove that the statement holds for all ...

Math - Use mathematical induction to prove that 5^(n) - 1 is divisible by four ...

Calculus - Use mathematical induction to prove that the statement holds for all ...

Math - Use mathematical induction to prove that 2^(3n) - 3^n is divisible by 5 ...

Pre-cal - use mathematical induction to prove that 1^2 + 2^2 + 3^2 + ... + n^2...

Discrete Math - Use mathematical induction to establish the following formula. ...

Math - Mathematical Induction - 3. Prove by induction that∑_(r=1)^n▒...

Calculus - Use mathematical induction to prove that each proposition is valid ...

Mathematical induction. I'm stuck. So far I have.. - For all integers n ≥ ...