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How would I prove the following by induction?$$1+3+3^2+3^3+\cdots+3^n=\frac{3^{n+1}-1}{2}$$ for all $n\geq 0.$

I kept trying to create a base case but I am not sure how many I need. I also seem to be carrying out the steps incorrectly.

If anyone knows how I can proceed with the following problem, I would greatly appreciate your assistance!

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closed as off-topic by Namaste, Leucippus, Simply Beautiful Art, JonMark Perry, Shailesh Oct 6 '17 at 1:05

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  • 1
    $\begingroup$ This is about as straightforward as induction proofs come. You say you need a base case; the problem asks to prove the equality for $n\ge0$, so the base case should be $n=0$. $\endgroup$ – Lord Shark the Unknown Oct 5 '17 at 5:05
  • $\begingroup$ You only need $1$ base case, so show the statement holds for $n=0$ (this is the first value in the range you are considering). Once you’ve shown that the statement is true for $n=0$, suppose it is true for SOME $k\geq0$ (you know $k$ exists because of the base case). Now show the statement is true for $k+1$ (be sure you know what the $k+1$ case looks like). $\endgroup$ – Clayton Oct 5 '17 at 5:08
  • $\begingroup$ "I kept trying to create a base case but I am not sure how many I need. I also seem to be carrying out the steps incorrectly." and this is why we highly recommend you include such context. As is, noone knows what steps you are doing, what mistakes you may be making, etc. $\endgroup$ – Simply Beautiful Art Oct 6 '17 at 0:42
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We start the induction with n=0.

It is $3^0=1=\frac{3^{0+1}-1}{2}=\frac{3-1}{2}=1\checkmark$.

Inductive claim:

For arbitrary $n\in\mathbb{N}$ holds $\sum_{k=0}^n 3^k=\frac{3^{n+1}-1}{2}$

Inductive step:

$n\mapsto n+1$

$\sum_{k=0}^{n+1} 3^k=\sum_{k=0}^{n} 3^k+3^{n+1}\stackrel{I.c}{=}\frac{3^{n+1}-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2\cdot 3^{n+1}}{2}=\frac{3\cdot 3^{n+1}-1}{2}=\frac{3^{n+2}-1}{2}\checkmark$

And we are done.

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  • $\begingroup$ The one seems to have slipped from the last stage. $\endgroup$ – Omry Oct 5 '17 at 5:12
  • $\begingroup$ Thank you, I edited it. :) $\endgroup$ – Cornman Oct 5 '17 at 5:13
  • $\begingroup$ @sktsasus: You can show similar, that $\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}$. You can do this exactly how I did it in my answer. You can also do this without any induction at all. $\endgroup$ – Cornman Oct 5 '17 at 5:15
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Induction step:

$1+3^1 +3^2+ 3^3+.....3^n+3^{n+1} =$

$\dfrac{3^{n+1} -1}{2} +3^{n+1}=$

$\dfrac{3^{n+1} +2×3^{n+1}-1}{2} =$

$\dfrac{(1+2)3^{n+1}-1}{2} =$

$\dfrac{3^{n+2}-1}{2}.$

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you can get you result like this :
$$ T=1+3+3^2+3^3+\cdots+3^n=S+1 \\ S=3+3^2+3^3+\cdots+3^n \\ {S \over 3} = 1+3+3^2+\cdots+3^{n-1} \\ S - {S \over 3}=3^n-1\\ S = (3^n-1)*3/2={{3^{n+1}-3} \over 2} $$ Result then is :
$$ T = {{3^{n+1}-3} \over 2}+1={{3^{n+1}-1} \over 2} $$

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  • $\begingroup$ You used the concept of geometric progression.OP asked for inductive approach. $\endgroup$ – tatan Oct 5 '17 at 5:43
  • $\begingroup$ Ok, sorry, i not read exactly his question $\endgroup$ – Daniel Pol Oct 5 '17 at 6:13
  • $\begingroup$ It's okay.. you can keep the answer but you would not be able to get upvotes maybe..;-) $\endgroup$ – tatan Oct 5 '17 at 6:55

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