Prove via induction $5^0 + 5^1 + 5^2 + \dots 5^n = \frac{5^{n+1}-1}{4}$

Question

Statement:

$$5^0 + 5^1 + 5^2 + \dots 5^n = \frac{5^{n+1}-1}{4}$$

I am having trouble prooving P(k+1) is true. Here is what I have so far:

$$\frac{5^{k+1} -1}{4} + 5^{k+1} = \frac{5^{k+2} -1}{4} $$

LHS $$ \textrm{ stuck here} = \frac{5^{k+1}}{4} - \frac{1}{4} + 5^{k+1} \\\\ OR \\ \textrm{ stuck after this } = \frac{5^{k+1} -1 + 4\cdot 5^{k+1}}{4}$$

No matter how I slice this equation, I am not able to get both sides equal. I have to ask is this even set up properly to begin with? What am I overlooking?

If you have $$\frac{5^{k+1} -1}{4} + 5^{k+1} = \frac{5^{k+2} -1}{4}$$ then you've got it. How did you get that, by the way? — Angina Seng, Mar 16 '19 at 19:51
I'm very confused by your work. You want to prove that $$\forall n\in \mathbb N\left(\sum \limits_{k=0}^n\left(5^k\right)=\dfrac{5^{n+1}-1}{4}\right)_.$$ You apparently are trying to start from the RHS and trying to reach the LHS. So you wrote $\dfrac{5^{k+2} -1}{4}=\dfrac{5^{k+1} -1}{4} + 5^{k+1}$. Now you just have to use the induction hypothesis on this. — Git Gud, Mar 16 '19 at 19:53
I finally got what you were trying to do. Let my confusion be of value to you. You were starting from your goal, which is $\displaystyle \sum \limits_{m=0}^{k+1}\left(5^m\right)=\dfrac{5^{k+2}-1}{4}$. You applied the induction hypothesis and transformed your goal (by means of equivalences) in $\dfrac{5^{k+1} -1}{4} + 5^{k+1} = \dfrac{5^{k+2} -1}{4}$. Your strategy consists of proving that your goal is equivalent to something obviously true (like $0=0$). You should make this more explicit in your writing. Lord Shark the Unknown was confused for the same reasons. — Git Gud, Mar 16 '19 at 20:02
Thanks Git Gud. I will definitely use this notation from now on, I just never learned how to express it formally — Evan Kim, Mar 16 '19 at 20:10

score 2 · Accepted Answer · answered Mar 16 '19 at 19:49

2

Hint: factor the numerator! I.e. what is $5^{k+1} + 4\cdot 5^{k+1}$?

answered Mar 16 '19 at 19:49

Minus One-Twelfth

7,413

Is it....$4\cdot 5^{k+1+1}$? This is where I get confused how to deal with it – Evan Kim Mar 16 '19 at 19:51
@EvanKim No; see my answer. – J.G. Mar 16 '19 at 19:52
Not quite. If you let $x = 5^{k+1}$, then it is $x+4x =, ???$. – Minus One-Twelfth Mar 16 '19 at 19:52
1

oh 5x. That makes more sense. It's weird that when I see $5^{k+1}+ 4\cdot 5^{k+1}$ I freeze but when I see $x + 4x$ it makes complete sense – Evan Kim Mar 16 '19 at 19:54

score 1 · Answer 2 · answered Mar 16 '19 at 19:50

1

First, it should be $n$, not $k$. Next factor out $5^{k+1}$ in $$5^{k+1}+4\cdot 5^{k+1}=\cdots\,?$$

answered Mar 16 '19 at 19:50

Bernard

175,478

Okay, I was factoring it out wrong, that's why I got stuck there – Evan Kim Mar 16 '19 at 19:56
2

That sort of things happens to all of us… – Bernard Mar 16 '19 at 19:57

score 1 · Answer 3 · answered Mar 16 '19 at 19:50

1

Multiplying by $4$ we get $$5^{k+1}-1+4\cdot 5^{k+1}=5^{k+1}-1$$ so it must be $$5^{k+1}+4\cdot 5^{k+1}=5^{k+1}$$ so $$5^{k+1}(1+4)=5^{k+2}$$

answered Mar 16 '19 at 19:50

Dr. Sonnhard Graubner

95,283

ahh, okay thank you. edit: can I ask why you added a -1 to the right hand side? Edit: Oh, it was from the other side of the equation, I get it now – Evan Kim Mar 16 '19 at 19:52

score 1 · Answer 4 · answered Mar 16 '19 at 19:51

1

$5^{k+1}+4\cdot 5^{k+1}=5\cdot 5^{k+1}=5^{k+2}$.

answered Mar 16 '19 at 19:51

J.G.

115,835

score 1 · Answer 5 · 2019-03-16T20:16:18.580

1

You are so close !

$$\frac{5^{k+1}-1+4\cdot5^{k+1}}{4} = \frac{5\cdot5^{k+1} -1}{4}.$$

edited Mar 16 '19 at 20:16

answered Mar 16 '19 at 19:53

Prove via induction $5^0 + 5^1 + 5^2 + \dots 5^n = \frac{5^{n+1}-1}{4}$

5 Answers5