Proving f(n) = 5 x 3^n - 4 with Mathematical Induction

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Homework Help Overview

The discussion revolves around proving a formula for a recurrence relation defined as f(n + 1) = 3f(n) + 8, with the initial condition f(1) = 11. Participants are exploring the validity of the proposed solution f(n) = 5 x 3^n - 4 through mathematical induction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants express uncertainty about how to begin the proof, while others suggest using mathematical induction by establishing a base case and assuming the formula holds for n=k before proving it for n=k+1. There is also a suggestion to consider an alternative approach by looking for a solution of a specific form based on the recurrence relation.

Discussion Status

The discussion is active, with participants sharing different perspectives on how to approach the problem. Some guidance has been offered regarding the induction process, and alternative methods are being considered, indicating a productive exploration of the topic.

Contextual Notes

There appears to be some confusion regarding the base case, as one participant mentions f(1)=1 instead of the stated f(1)=11. This discrepancy may affect the assumptions being discussed.

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Homework Statement


Let f(n + 1) = 3f(n) + 8, with f(1) = 11. Prove by induction that f(n) = 5 x 3^n - 4.

Homework Equations


The Attempt at a Solution


I don't even know where to start! Any help would be appreciated. Thanks. :-)
 
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Well since it gives you the base case, f(1)=1, you can show it's true for the base case n=0.

Now you assume it is true for n=k, that is,
[tex]3f(k)+8=5.3^k-4[/tex]

Now prove it true for n=k+1,

[tex]3f(k+1)+8=5.3^{k+1}-4[/tex]
 
You don't really need induction, the recurrence relation is given by:
[tex] a_{n+1}=3a_{n}+8[/tex]
This suggests you look for a solution of the form:
[tex] a_{n}=\alpha\cdot 3^{n}+\beta[/tex]
For some [tex]\alpha ,\beta[/tex], then it is just a matter of plugging this into the equation to obtain values for [tex]\alpha[/tex] and [tex]\beta[/tex].

Mat
 
Mentallic said:
Well since it gives you the base case, f(1)=1, you can show it's true for the base case n=0.

Now you assume it is true for n=k, that is,
[tex]3f(k)+8=5.3^k-4[/tex]

Now prove it true for n=k+1,

[tex]3f(k+1)+8=5.3^{k+1}-4[/tex]

Nice one, thanks!
 

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