Induction: 1+2+2^2+2^3+ +2^n-1=2^n-1

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Homework Help Overview

The problem involves using mathematical induction to prove the equality \(1 + 2 + 2^2 + 2^3 + \ldots + 2^{n-1} = 2^n - 1\) for all integers \(n \geq 1\). The discussion centers around the setup and execution of the induction proof.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the base case for \(n=1\) and the inductive step from \(k\) to \(k+1\). There is confusion regarding the correct formulation of the terms in the equation, particularly the placement of parentheses and the interpretation of the summation. Some participants question the validity of the expressions used and suggest clarifying the notation.

Discussion Status

Multiple interpretations of the problem are being explored, with participants providing feedback on the original poster's attempts. There is an acknowledgment of confusion regarding the expressions used, and some guidance has been offered to clarify the notation and approach to the proof.

Contextual Notes

Participants note that the original statement may contain errors in notation, specifically regarding the placement of terms and the use of parentheses, which could lead to misunderstandings in the proof process.

Daaniyaal
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1. Homework Statement [/b]
Use mathematical induction to prove the following statements are true for all integers n≥1
1+2+2^2+2^3+...+2^n-1=2^n-1



Attempt at a solution:


For n=1

1+2^(1-1)=2^1-1
1=1
∴ it is true.


Let Sn=Sk
If it is true for k, it must also be true for k+1

1+2+2^2+2^3+...+2^k+2^k+1-1=(2^k+1)-1

This is the part I have a slight confusion at, I keep getting something left over and the sides don't equal each other :(
 
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Daaniyaal said:
1. Homework Statement [/b]
Use mathematical induction to prove the following statements are true for all integers n≥1
1+2+2^2+2^3+...+2^n-1=2^n-1



Attempt at a solution:


For n=1

1+2^(1-1)=2^1-1
1=1
∴ it is true.


Let Sn=Sk
If it is true for k, it must also be true for k+1

1+2+2^2+2^3+...+2^k+2^k+1-1=(2^k+1)-1

This is the part I have a slight confusion at, I keep getting something left over and the sides don't equal each other :(

Use brackets: you have written
1+2+2^2+2^3+...+2^n-1=2^n-1 , which is plainly impossible, because it would require that we have 1+2+2^2+2^3+... = 0 (in order to have 2^n-1=2^n-1 on both sides). If you mean 2^(n-1), then that is what you need to say.

RGV
 
You want to prove that if, for some k, [itex]2^0+ 2^1+ \cdot\cdot\cdot+ 2^{k-1}= 2^k-1[/itex] then [itex]2^0+ 2^1+ \cdot\cdot\cdot+ 2^{k-1}+ 2^k= 2^{k+1}-1[/itex].
(What you wrote, 1+ 2^1+ ...+ 2^n-1= 2^n-1 is, as Ray Vickson said, clearly impossible because you have "2^n- 1" on both sides but with additional positive terms on the left. Of course, you meant 2^(n-1) on the left and (2^n)- 1 on the right. Those are very different and you can't ask people to guess what you mean.)

To go from k to k+1, obviously you are adding [itex]2^k[/itex] to the sum on the left so do it on the right:
[tex]2^0+ 2^1+ \cdot\cdot\cdot+ 2^{k-1}+ 2^k= (2^k- 1)+ 2^k[/tex]

What do you get when you combine those two "[itex]2^k[/itex]" terms on the right?
 
Last edited by a moderator:
I figured it out, thanks!
 

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