Proof By Induction Help?

Question

I've been working through proof by induction and i'm stuck on this question. Can somebody provide some help? $$ 2^n-1=\sum_{i=0}^{n-1}2^i\text{ for }n\ge 1$$

Which step in the induction process are you stuck on? What have you attempted so far? What is your understanding of induction? — Mufasa, Jan 20 '15 at 22:09
If you learn some telescopy such inductive proofs become very trivial. — Bill Dubuque, Jan 20 '15 at 22:12
http://math.stackexchange.com/questions/761156/proof-by-induction-that-sum-j-0n-2j-2n1-1 http://math.stackexchange.com/questions/209518/how-to-prove-by-induction-that-sumn-i-12i-1-2n-1 http://math.stackexchange.com/questions/22599/how-do-i-prove-this-by-induction-sum-of-powers-of-2 http://math.stackexchange.com/questions/180169/summation-equation-for-2x-1 ... and many other question you can find on this site. (For example, looked at related and linked questions at those question I've mentioned here.) — Martin Sleziak, Feb 19 '15 at 09:33

barak manos · Answer 1 · 2015-01-20T23:05:27.283

2

First, show that this is true for $n=1$:

$2^1-1=\sum\limits_{i=0}^{1-1}2^i$

Second, assume that this is true for $n$:

$2^n-1=\sum\limits_{i=0}^{n-1}2^i$

Third, prove that this is true for $n+1$:

$2^{n+1}-1=2\cdot2^n-1$
$2\cdot2^n-1=2\cdot2^n-2+1$
$2\cdot2^n-2+1=2\cdot(2^n-1)+1$
$2\cdot(2^n-1)+1=1+2\cdot(2^n-1)$
$1+2\cdot(2^n-1)=1+2\cdot\sum\limits_{i=0}^{n-1}2^i$ assumption used here
$1+2\cdot\sum\limits_{i=0}^{n-1}2^i=1+\sum\limits_{i=1}^{n}2^i$
$1+\sum\limits_{i=1}^{n}2^i=2^0+\sum\limits_{i=1}^{n}2^i$
$2^0+\sum\limits_{i=1}^{n}2^i=\sum\limits_{i=0}^{n}2^i$

edited Jan 20 '15 at 23:05

answered Jan 20 '15 at 22:26

barak manos

43,599

thank you i've got it - but can i just ask how you got to the third part bullet point 1? - how did you get to 2(2n - 1) – Mathematica Jan 20 '15 at 22:39
@Mathematica: Which bullet do you mean, the first one in the third part? – barak manos Jan 20 '15 at 22:40
@Mathematica: I just tried to represent the proof subject ($2^{n+1}-1$) using the assumption subject ($2^{n}-1$). The ratio between them is approximately $2$, right? So I multiplied the assumption subject by $2$, and then added what was necessary in order to make them equal (i.e., $+1$). – barak manos Jan 20 '15 at 22:44
@Mathematica: I've slightly expanded it in order to make it easier for you to understand. – barak manos Jan 20 '15 at 22:48
@Mathematica: I've significantly expanded it in order to make it easier for you to understand. – barak manos Jan 20 '15 at 22:54
thank you so much for the help - I'm struggling to understand one thing - how did you get [2 * ] (2n -1) on the first line? – Mathematica Jan 20 '15 at 23:07
@Mathematica: You're welcome. Are you struggling to understand $2^{n+1}=2\cdot2^n$??? – barak manos Jan 20 '15 at 23:12
yes i don't understand how you got the 2 * – Mathematica Jan 20 '15 at 23:18
@Mathematica: Let me get this right - you don't understand that, but you understand all the rest??? – barak manos Jan 20 '15 at 23:26
Yeah - I've understood the basic concept of whats going on but i want to confirm how you got that – Mathematica Jan 20 '15 at 23:32
@Mathematica: Do you understand what $2^n$ means? – barak manos Jan 20 '15 at 23:32
yes I understand that - are you just using the indices rule? – Mathematica Jan 20 '15 at 23:33
@Mathematica: Then surely you understand what happens when you multiply it by $2$... right? – barak manos Jan 20 '15 at 23:35
1

@barakmanos oh my god, this is the funniest comment section ever. I laughed so hard when you were shocked by him not understanding $2^{n+1} = (2)(2^n)$ – elbarto Apr 15 '15 at 20:26
1

@elbarto: I did do my best trying to explain it though, didn't I? – barak manos Apr 17 '15 at 09:36

score 0 · Answer 2 · answered Jan 20 '15 at 22:10

0

Hint : $2^{n+1}-1=2(2^n-1)+1$.

answered Jan 20 '15 at 22:10

Guest

4,268

score 0 · Answer 3 · answered Jan 20 '15 at 22:19

Guest has given a very good hint. Alternatively, you can generalize:

For any $x,y \in \Bbb{R}$ then $$x^n-y^n = (x-y)\sum_{i=0}^{n-1}y^ix^{n-1-i}$$ If you prove this by induction, then your proof follows as a special case of $x=2, y=1$. Your original proof is probably easier than this one, but this one is a good result to have regardless.

score 0 · Answer 4 · answered Jan 20 '15 at 22:25

Although the hint by Guest is quite good, as graydad pointed out, I imagine you are probably struggling with some summation manipulation (or am I wrong?). Here's a brief sketch (I imagine you can fill in all of the fine details and make the argument flow like a written proof should): \begin{align} 2^{k+1}-1 &= 2\cdot(2^k-1)+1\\[1em] &= 2\cdot\sum_{i=0}^{k-1}2^i + 1\\[1em] &= \sum_{i=1}^k2^i + 1\\[1em] &= \sum_{i=0}^k2^i. \end{align} Does that make more sense now?

Proof By Induction Help?

4 Answers4