Simplify 2^k -1 + 2^[ (k+1) -1] ? In trying to prove ... + 2^(n-1) = 2^n -1.

[jddoxtator]

In trying to prove ... + 2^(n-1) = 2^n -1.
To do this I substituted n for k and tried to prove for ( k + 1 ).
This leads to ... + 2^( k - 1 ) + 2^[( k + 1 ) - 1 ] = 2^k - 1 +2^[( k+1 ) - 1 ]
In order to prove this I have to simplify = 2^k -1 + 2^[( k + 1 ) -1 ].

To my knowledge, the only way to do this multiply exponents of the same base as different exponents of the same base cannot be added.
The only other way would be to isolate the same bases on either side of the equals sign, but then there is no way to deal with base - 1.
Not sure how to proceed.
The best I have done is = 4^k -1 which could be = 2(2^k) - 1 or = 2^( k + 1 ) - 1

 

[Dr.Peterson]

In trying to prove ... + 2^(n-1) = 2^n -1.

Please clarify what you are trying to prove. Clearly it is not "2^(n-1) = 2^n -1", which is not true.

Is it a summation, and you are showing only the last term? Perhaps an image of the problem in print or handwritten will help.

The best I have done is = 4^k -1 which could be = 2(2^k) - 1 or = 2^( k + 1 ) - 1

No, you can't change the base; 4^k is not equal to 2(2^k). But your last expression is a correct simplification of the expression in the title, and is probably exactly what you need if you are trying to prove a summation by induction.

 

[jddoxtator]

Please clarify what you are trying to prove. Clearly it is not "2^(n-1) = 2^n -1", which is not true.

Is it a summation, and you are showing only the last term? Perhaps an image of the problem in print or handwritten will help.


No, you can't change the base; 4^k is not equal to 2(2^k). But your last expression is a correct simplification of the expression in the title, and is probably exactly what you need if you are trying to prove a summation by induction.

Yes, it is summation by induction.
Guess I should have added a few more terms instead of just "..."
I am having a real hard time with these summation proofs because the ( k + 1 ) seems to be handled like a variable identity instead of a binomial with typical algebra factoring rules.
It seems in every case you are factoring with the goal of ( k + 1 ) instead of fully factored.
It is a weird way of thinking as you are going forward and then in reverse in your factoring to reveal the original equation with the ( k + 1 ) as the variable.

Most difficult case I came across was the Fibonacci sequence represented as f1 + f2 + ... + fn. Since the first two terms are identical, the n=1 of the equation to prove was f1 = f2, which technically is true by the range of a Fibonacci sequence and was then a factor in the proof where the identity of f2 had to be swapped out to f1 to complete the proof. Mind boggling...

 

[khansaheb]

Yes, it is summation by induction.
Guess I should have added a few more terms instead of just "..."
I am having a real hard time with these summation proofs because the ( k + 1 ) seems to be handled like a variable identity instead of a binomial with typical algebra factoring rules.
It seems in every case you are factoring with the goal of ( k + 1 ) instead of fully factored.
It is a weird way of thinking as you are going forward and then in reverse in your factoring to reveal the original equation with the ( k + 1 ) as the variable.

Most difficult case I came across was the Fibonacci sequence represented as f1 + f2 + ... + fn. Since the first two terms are identical, the n=1 of the equation to prove was f1 = f2, which technically is true by the range of a Fibonacci sequence and was then a factor in the proof where the identity of f2 had to be swapped out to f1 to complete the proof. Mind boggling...

Please post the COMPLETE problem EXACTly as it was presented to you. A photocopy/picture will be useful.

 

[Dr.Peterson]

I am having a real hard time with these summation proofs because the ( k + 1 ) seems to be handled like a variable identity instead of a binomial with typical algebra factoring rules.

It's ordinary algebra, but you do have to get used to an unfamiliar sort of goal.

But we can't really help without seeing (a) the actual problem you are working on (including the first terms, and the condition on k), and (b) your actual work, from the start (not just this one step).

Most difficult case I came across was the Fibonacci sequence represented as f1 + f2 + ... + fn. Since the first two terms are identical, the n=1 of the equation to prove was f1 = f2, which technically is true by the range of a Fibonacci sequence and was then a factor in the proof where the identity of f2 had to be swapped out to f1 to complete the proof. Mind boggling...

If you want to discuss that example, again, please show the entire work you are confused by, and point out the details you don't understand. The more you show, the better. (What you say here makes no sense without context, and is probably a misunderstanding we can point out when we see it all.)

But that should be in a separate thread, to avoid confusion.

 

[MathNugget]

In trying to prove ... + 2^(n-1) = 2^n -1.
To do this I substituted n for k and tried to prove for ( k + 1 ).
This leads to ... + 2^( k - 1 ) + 2^[( k + 1 ) - 1 ] = 2^k - 1 +2^[( k+1 ) - 1 ]
In order to prove this I have to simplify = 2^k -1 + 2^[( k + 1 ) -1 ].

To my knowledge, the only way to do this multiply exponents of the same base as different exponents of the same base cannot be added.
The only other way would be to isolate the same bases on either side of the equals sign, but then there is no way to deal with base - 1.
Not sure how to proceed.
The best I have done is = 4^k -1 which could be = 2(2^k) - 1 or = 2^( k + 1 ) - 1

Attempting to guess this from the little info...
Are you asking how to prove [imath]1+2+...+2^{n-1}=2^n-1[/imath] through induction? Because it quite looks like it... from the whole thing about n and k...

 

[MathNugget]

I gave the thing a 2nd look (now it seems rather obvious what your problem was): you made a mistake there, that's why it's not working and the "pattern" breaks:

This leads to ... + 2^( k - 1 ) + 2^[( k + 1 ) - 1 ] = 2^k - 1 +2^[( k+1 ) - 1 ]

hint: why would the exponent suddenly jump from k-1 to k+1?

If you write the steps carefully, you'd notice P(k) looks like this:
[math]P(k): 1+2+...+2^{k-1}=2^k-1[/math]So P(k+1) would not be [imath]P(k)+2^{k+1}[/imath]. I suppose that's why you couldn't get the result... You have to think again about who P(k+1) is.