1

Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$

The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?

Ivo Terek
  • 80,301
gger234
  • 305
  • 2
  • 5
  • 16
  • Is this really algebra-precalculus? – Shahar Aug 27 '14 at 20:27
  • See the second answer here: http://math.stackexchange.com/questions/22599/how-do-i-prove-this-by-induction-sum-of-powers-of-2/ – user133281 Aug 27 '14 at 20:27
  • @Shahar Well, it's not calculus! But yes, I guess it'd appear in algebra $2$ in high school. –  Aug 27 '14 at 20:27
  • @SanathDevalapurkar Not really calculus but I didn't learn this in algebra either. What should I tag this? – Shahar Aug 27 '14 at 20:28
  • alright, so I guess I get 3^(k-1)+1) which is 3^k? @user133281 – gger234 Aug 27 '14 at 20:29
  • @Shahar I believe there's an induction tag. Oh wait, Ivo Terek already edited it. Never mind! –  Aug 27 '14 at 20:30
  • you equality is wrong !! For $n=0$, $n=1$... try, it doesn't work. – idm Aug 27 '14 at 20:31
  • @Shahar : Nothing is "algebra-precalculus". The tag should be abolished. Tags should be about subject matter, not about where something is located in a curriculum. Curricula are evil. – Michael Hardy Aug 27 '14 at 20:35
  • @Shahar : I think of this as a standard part of high-school math that is covered before one gets to calculus, but in in "our" present "system" (ok, maybe it is in some respects a system) that means 99% of those who've gone through that course will never have heard of it. – Michael Hardy Aug 27 '14 at 20:37
  • @MichaelHardy What about stuff like $\cos(x)+3=2$? – Shahar Aug 27 '14 at 20:37
  • What about it? ${}\qquad{}$ – Michael Hardy Aug 27 '14 at 20:38
  • @MichaelHardy Is that under the Common Core or something? I never learned this in algebra class. – Shahar Aug 27 '14 at 20:39
  • @MichaelHardy I meant what would you classify a question like that, it incorporates both algebra and precalculus. – Shahar Aug 27 '14 at 20:40
  • @Shahar : I have no idea what the content of "common core" is. I have the impression it's about prescribing that all humans should learn the same things at the same ages. That is evil. – Michael Hardy Aug 27 '14 at 20:46
  • @MichaelHardy All students of the United States, if that makes it less evil ^_^. – Shahar Aug 28 '14 at 00:58

4 Answers4

7

  1. Show $3^{1-1}=\dfrac{3^1-1}{2}$

  2. Assume $\sum\limits_{k=0}^{n-1}3^k=\dfrac{3^n-1}{2}$

  3. Prove $\sum\limits_{k=0}^{n}3^k=\dfrac{3^{n+1}-1}{2}$:

    • $\sum\limits_{k=0}^{n}3^k=(\sum\limits_{k=0}^{n-1}3^k)+3^n$

    • $(\sum\limits_{k=0}^{n-1}3^k)+3^n=\dfrac{3^n-1}{2}+3^n$

    • $\dfrac{3^n-1}{2}+3^n=\dfrac{3^n-1+2\cdot3^n}{2}$

    • $\dfrac{3^n-1+2\cdot3^n}{2}=\dfrac{3^n+2\cdot3^n-1}{2}$

    • $\dfrac{3^n+2\cdot3^n-1}{2}=\dfrac{3\cdot3^n-1}{2}$

    • $\dfrac{3\cdot3^n-1}{2}=\dfrac{3^{n+1}-1}{2}$

Note that the induction-step is applied only at the second bullet.

barak manos
  • 43,599
2

Suppose that $x \neq 1$. We wish to show by inducting on $n$ that $\sum\limits_{k=0}^{n-1}x^{k} = \frac{x^{n} - 1}{x-1}$.

Then $1 = \frac{x-1}{x-1}$ so that the formula holds for $n = 1$. Suppose that we have $\sum\limits_{k=0}^{n-1}x^{k} = \frac{x^{n} - 1}{x-1}$. We must show that $\sum\limits_{k=0}^{n}x^{k} = \frac{x^{n+1} - 1}{x-1}$.

We have $\sum\limits_{k=0}^{n}x^{k} = \sum\limits_{k=0}^{n- 1}x^{k} + x^{n} = \frac{x^{n} - 1}{x-1} + x^{n}\frac{x -1}{x-1} = \frac{x^{n} - 1 + x^{n+1} - x^{n}}{x - 1} = \frac{x^{n+1} - 1}{x-1}$

user84413
  • 27,747
Daniel Akech Thiong
  • 4,129
0

For n+1 it is

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

$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{3\cdot 3^{n}-1}{2}$

$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{(1+2)\cdot 3^{n}-1}{2}$

$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{1\cdot 3^{n}-1}{2}+\frac{2\cdot 3^{n}}{2}$

$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\left( \frac{1\cdot 3^{n}-1}{2}\right) + 3^n$

callculus42
  • 31,012
0

For the base case simply show that the premise is true for the smallest positive integer: $\mathsf P(1)$.

For the iterative case, you have to show that if the premise holds for $n$, then it is implied to hold for $n+1$.   $\mathsf P(n)\implies\mathsf P(n+1)$

Do that by substituting $n+1$ for $n$ and demonstrate that the left and right hand side both contain terms of the original premise and the same modifier.

Here we show that modifier is the addition of $3^n$.

$\begin{align} \text{Prove }\forall n\in \mathbb Z^+: \sum_{k=0}^{n-1} 3^k & = \frac {3^n-1}{2} & \text{premise } \\[3ex] 3^0 & = \frac{3^1-1}{2} & \text{base case} \\[3ex] 3^n + \sum_{k=0}^{n-1} 3^k & = \sum_{k=0}^{n} 3^k \\[1ex] 3^n+\frac {3^n-1}{2} & = \frac{3\cdot 3^n-1}{2} \\[2ex] \therefore \sum_{k=0}^{n-1} 3^k & = \frac {3^n-1}{2} \implies \sum_{k=0}^{n} 3^k = \frac{3^{n+1}-1}{2} &\text{iterative step } \\[2ex]\therefore \forall n\in\mathbb Z^+: \sum_{k=0}^{n-1} 3^k & = \frac {3^n-1}{2} & \text{by induction} \end{align}$

Graham Kemp
  • 133,231