Solve Problem with Proof by Induction

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    Induction Proof
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Homework Help Overview

The discussion revolves around using mathematical induction to prove a formula related to a geometric series involving powers of 5. The original poster presents a summation that needs to be proven for all integers n greater than or equal to zero.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the steps involved in proof by induction, including establishing a base case and formulating an inductive hypothesis. There are questions about the clarity of the original poster's steps and the need for a more detailed explanation of their approach.

Discussion Status

Some participants have provided guidance on how to structure the proof, particularly emphasizing the importance of the inductive step. There is an ongoing exploration of the summation and its components, but no consensus has been reached on the complete proof.

Contextual Notes

One participant notes the potential risks of uploading documents, suggesting that the original poster should share their work directly in the thread. The original poster's steps appear to be incomplete, and there is a request for clarification on their reasoning.

smh745
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I need help to solve this problem

Use induction to prove that, for n>=0:

3*5^0 + 3*5^1 + 3*5^2 + 3*5^3 + ...+ 3*5^n = 3*(5^(n+1)-1)/4

in other word


n

[tex]\sum[/tex] 3*5 k= 3*(5 n+1-1) / 4
k= 0
 
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Assume true for n=N and now prove true for n=N+1.
Do you know how to do a proof by induction?
 
I did these steps and I tried to complete the rest but I don't know some of the steps

and what I is in the attached doc.
 

Attachments

Try not to upload word documents as these usually contain viruses. Can you type out the steps you did?
 
Basis: n= 0
0
[tex]\sum[/tex]3*5^ 0 = 3

k= 0




3*(5 0+1-1) / 4 = 3




Assume:

n

[tex]\sum[/tex] 3*5^k= 3*(5 n+1-1) / 4
k= 0



Prove:

n+1

[tex]\sum[/tex] 3*5 ^k= 3*(5 (n+1)-1) / 4
k= 0


________________________________________
Proof:

n+1
[tex]\sum[/tex] 3*5^ k =

k= 0
 
As you have the base case, think about the sum:

3*5^0 + 3*5^1 + 3*5^2 + 3*5^3 + ...+ 3*5^n + 3*5^(n+1)

which is now your inductive step, as rock.freak667 suggested.

What are the two ways in which this can also be written with the information you already have?

The Bob
 

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