Solve Complex Function: Step-by-Step Guide

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Homework Help Overview

The discussion revolves around a complex function problem involving the equation f(n) = 15^n + 8^{n-2}. Participants are exploring the manipulation of this function to express f(n+1) - 8f(n) as a multiple of 15^n.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks clarification on the initial steps of a provided solution and requests resources for understanding the topic. Other participants engage in algebraic manipulation of the function and discuss the cancellation of terms in their calculations.

Discussion Status

The conversation is ongoing, with participants sharing insights and calculations. There is an indication of productive exploration as one participant attempts to clarify their understanding of the problem, while others contribute to the algebraic reasoning involved.

Contextual Notes

One participant noted the absence of an attachment initially, which may have limited the context for the discussion. The original poster has expressed uncertainty about the steps in the solution, indicating a need for further clarification on the methods used.

trojsi
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Hi,
please find attached the problem and the answer. I can't anderstand the first and second step of the answer. I only need a link for the topic or maybe method the used. Bdw I haven't tried to figure out the second answer but for now I only need help tackling the first one.
 

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thanks for telling.. I uploaded it now :)
 
I have the feeling I have done this before!

You have the equation [itex]f(n)= 15^n+ 8^{n-2}[/itex] and want to write f(n+1)- 8 f(n) as a multiple of [itex]15^n[/itex].

Okay, [itex]f(n+1)= 15^{n+1}+ 8^{n+1- 2}[/itex][itex]= 15^{n+1}- 8^{n-1}[/itex] and [itex]8f(n)= 8(15^n)- 8^{n-2})=[/itex][itex]8(15^n)- 8(8^{n-2})=[/itex][itex]8(15^n)- 8^{n-1}[/itex]

Now, when you subtract the two "[itex]8^{n-1}[/itex]" terms cancel and you get [itex]15^{n+1}- 8(15^n)= 15(15^n)- 8(15^n)= (15- 8)15^{n+1}[/itex].
 

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