Induction Proof Help: Understanding Equations for Homework | Prof. Note's Tips

[Kingyou123]

Homework Statement

Homework Equations

Prof. Note's.

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

 

[HallsofIvy]

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.

 

[ehild]

Homework Statement

View attachment 95427

Homework Equations

Prof. Note's. View attachment 95431

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
View attachment 95428

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

 

[Kingyou123]

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

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

 

[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.

 

[HallsofIvy]

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

 

[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?

 

[Kingyou123]

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.

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

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

 

[TheMathNoob]

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.

Homework Statement

View attachment 95427

Homework Equations

Prof. Note's. View attachment 95431

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
View attachment 95428

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