1. ## Proof by Induction

Prove 24|2*7^n+3*5^n-5

I showed true for n=1:
2*7+3*5-5=14+15-5=24
Thus divisible by 24
Assume true for all k>=1. Need to prove true for n=k+1
Then 2*7^(k+1)+3*5^(k+1)-5=14*7^k+15*5^k-5
I get this far and now I get stuck simplifying enough to show it's true.

2. Originally Posted by kathrynmath
Prove 24|2*7^n+3*5^n-5

I showed true for n=1:
2*7+3*5-5=14+15-5=24
Thus divisible by 24
Assume true for all k>=1. Need to prove true for n=k+1
Then 2*7^(k+1)+3*5^(k+1)-5=14*7^k+15*5^k-5
I get this far and now I get stuck simplifying enough to show it's true.
$\displaystyle 14\cdot7^k+15\cdot5^k-5 = 12\cdot\left(7^k+5^k\right)+\left(2\cdot7^k+3\cdot 5^k-5\right)$

Now note that $\displaystyle 7^k+5^k$ is even.

3. Originally Posted by kathrynmath
Prove 24|2*7^n+3*5^n-5

I showed true for n=1:
2*7+3*5-5=14+15-5=24
Thus divisible by 24
Assume true for all k>=1. Need to prove true for n=k+1
Then 2*7^(k+1)+3*5^(k+1)-5=14*7^k+15*5^k-5
I get this far and now I get stuck simplifying enough to show it's true.

$\displaystyle 2\cdot 7^{k+1}+3\cdot 5^{k+1}-5=[2\cdot 7^k+3\cdot 5^k-5]+12[7^k+5^k]$ ,so the first parentheses

above is divisible by 24 by the inducive hypothesis, and the second parentheses is obviously even, so...

Tonio