# Induction Proof Help...

1. Feb 6, 2016

### Kingyou123

1. The problem statement, all variables and given/known data

2. Relevant equations
Prof. Note's.
3. The attempt at a solution
I'm on the 3 line where my Prof. combines both equations, I'm confused on what my equation should look. Her's was (n+1)(n+1)+1)/2

2. Feb 6, 2016

### HallsofIvy

Staff Emeritus
Saying that "4 divides $5^n- 1$ is NOT just a "reference" to $\frac{5^n- 1}{4}$! It is the statement that $\frac{5^n- 1}{4}$ is an integer. That is, $\frac{5^n- 1}{4}= k$ for some integer k. Further the "n+1" form of the formula is not $\frac{5^{n+1}- 1}{4}+ (n+1)$. I don't where you got that additional "(n+1)"! Replacing n by n+1 in $\frac{5^{n}- 1}{4}$ is just $\frac{5^{n+1}- 1}{4}$.

Now, of course, you want to "algebraically" go back to the "$5^n$" and to do that use the fact that $5^{n+1}= 5(5^n)$.
It will be helpful to use $5(5^n)- 1= 5(5^n)- 5+ 4$.

3. Feb 6, 2016

### ehild

You can write 5n+1 as 5 * 5n, and 5n+1 - 1 = 5 * 5n -5 +4 = 5(5n-1) + 4.

4. Feb 6, 2016

### Kingyou123

Sorry, I just noticed this but should my equation be 4 l 5^n-1 or is what have okay?

5. Feb 6, 2016

### ehild

Yes. As 5n-1 is divisible by 4 , the first therm of 5(5n-1) + 4 is divisible by 4, and the second term is just 4.

6. Feb 6, 2016

### HallsofIvy

Staff Emeritus
"4 l 5^n-1" is NOT even an equation!

7. Feb 6, 2016

### Kingyou123

Okay sorry, I'm a bit confused now. So (5^(n)-1)/4 is correct, right ? And would I follow what my prof. did, so I set my work for n+1 to 5(5n-1) + 4 or is that the induction step?

8. Feb 6, 2016

### Kingyou123

Sorry, I just refreshed my page and your comment appeared, thank you for the help :)

9. Feb 15, 2016

### TheMathNoob

I think that you have to go from the fact that if (5^n)-1 is divisible by 4 then (5^n)-1=4k where k is a constant. Now, how can you apply this to 5((5^n)-1)+4?. Think of substitution

Last edited: Feb 15, 2016