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

Basis case $n= 0$:

$1^0 = 1 \;\;\;\;\;\;\;\;\;\;\;\; \frac{5^{1+1}-1}{4}=1$

Assume true for $n=k$: $$1 + 5 + 5^2 + \ldots + 5^k = \frac{5^{k+1}-1}{4}$$

Need to show for $n=k+1$: $$1 + 5 + 5^2 + \ldots + 5^{k+1} = \frac{5^{k+1+1}-1}{4} = \frac{5^{k+2}-1}{4}$$

Induction proof $$\frac{5^{k+1}-1}{4} + 5^{k+1}=\frac{5^{k+1}-1}{4} + \frac{4(5^{k+1})}{4}$$ This is where I am stuck and do not know if I am even right with this at all.

jdc987
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  • Just put that last term over the common denominator $4$, then factor $5^{k+1}$ out of those two terms and what do you get? – Rookatu Mar 25 '14 at 00:48
  • Something that might help. There is a typo in the last expression after "need to show". It should be $5^{k+2}$. Given $5^{k+1}$, how can you get to $5^{k+2}$? – Henry Swanson Mar 25 '14 at 00:54
  • @HenrySwanson Thanks for that, I fixed it above. – jdc987 Mar 25 '14 at 00:57
  • $5ˆ({k+1})+4(5ˆ({k+1}))=5(5ˆ({k+1}))=5ˆ({k+2})$, that is what you want. – fmoura2005 Mar 25 '14 at 00:57
  • @fmoura2005 What happened to the 4 on the bottom and the -1? – jdc987 Mar 25 '14 at 01:00
  • The final result is $({5^{k+2}-1})/4$, proving the inductive step. – fmoura2005 Mar 25 '14 at 01:04
  • @jdc987 They just stay where they are. Not every step will involve every component of the expression, so he just left out the unimportant ones. (You should write them in though) – Henry Swanson Mar 25 '14 at 01:05
  • @HenrySwanson I understand that, but what I don't understand is why the 4 on the bottom is gone and why the -1 is gone. Basically, how he got to that point is where i'm lost. – jdc987 Mar 25 '14 at 01:10
  • Less effort? An assumption that you could figure the rest out? I'll write it out in full, so you can compare the two. $\frac{5^{k+1} - 1}{4} + \frac{4 \cdot 5^{k+1}}{4} = \frac{5^{k+1} - 1 + 4 \cdot 5^{k+1}}{4} = \frac{5 \cdot 5^{k+1} - 1}{4} = \frac{5^{k+2} - 1}{4}$. The $4$ and $-1$ aren't really gone. But imagine you were explaining the process to a friend. You wouldn't explain it as "first you take [long string of words] and turn it into [slightly different string], then...". You'd describe the most important part: "you combine the fractions, and group the $5^{k+1}$ terms together". – Henry Swanson Mar 25 '14 at 01:13
  • @HenrySwanson Thanks for writing it out. I'm having a hard time understanding these proofs so I really had no idea how he arrived at that step. Why does the 4 get removed after the second equal sign? – jdc987 Mar 25 '14 at 01:16
  • It's because you're adding $1$ of a thing to $4$ of that thing. It doesn't really matter that the thing happens to be $5^{k+1}$, it all boils down to the old $1$ apple + $4$ apple kind of stuff. (If you want to be formal about it, it's the distributive law: $x + 4x = 1x + 4x = (1+4)x = 5x$) – Henry Swanson Mar 25 '14 at 01:21

1 Answers1

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This is a geometric series, with common ratio = 5.

$$1 + q + q^2 + ... + q^n = (1 - q^{n+1})/(1-q)$$

Proof of this assumption by induction :

The case n=0 is trivial ($ 1 = 1 $)

Assume you have the previous result for n.

$$1+q+...+q^{n+1} = (1-q^{n+1})/(1-q) + q^{n+1}$$

$$1+...+q^{n+1} =(1-q^{n+1} + q^{n+1}\cdot (1-q) ) / (1-q)$$

If you develop, you get the Right hand side :

$$ (1 - q^{n+1} + q^{n+1} - q^{n+2})/(1-q)$$

Thus,

$$(1-q^{n+2})/(1-q)$$

Which is the expected result for n+1

T_O
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  • Sorry, that was a typo. – jdc987 Mar 25 '14 at 00:41
  • Ok, so if you don't know this result about geometric series, back to proving by induction the general result ! – T_O Mar 25 '14 at 00:42
  • @jdc987 I edited my post, this should be better. You can apply for q=5 or simply learn the formula and apply it in your case. If you do need a proof by induction, my proof should be enough to help you get yours ;) – T_O Mar 25 '14 at 00:47
  • @jdc987 by the way if you are stuck with your last line, just replace the 4 in $4(5^{k+1})$ by (5-1) and develop. You should find the result (it's the exact same thing as my proof really) – T_O Mar 25 '14 at 00:54