What is the solution to r^(n+1)*(1-r) in calculus?

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Homework Help Overview

The discussion revolves around the expression r^(n+1)*(1-r) in the context of a proof involving geometric series in calculus. Participants are exploring the implications of this expression and its relation to the formula for the sum of a geometric series.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the expression and its relevance to proving a formula for the sum of a geometric series. Questions arise about the base case in induction, the validity of n being zero, and the implications of different values of n.

Discussion Status

The discussion is active, with participants providing insights into the base case of the induction proof and questioning the assumptions about the variable n. Some participants suggest alternative ways to approach the proof without induction, while others clarify the meaning of the left-hand side of the equation when n equals zero.

Contextual Notes

There is a focus on the conditions under which the formula holds, particularly the requirement that r does not equal 1 and the implications of n being greater than or equal to zero. Participants are also considering the consequences of different interpretations of the series notation.

Roscoe1989
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1. can someone tell me what r^(n+1)*(1-r)=? I need it to solve a proof, but my math is so rusty! thanks!



2. r^(n+1)*(1-r)
 
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What is the question...? Like the real question.
 
prove by induction on n that
1+r+r^2...+r^=1-r^n+1/1-r
if r doesn't equal 1
 
Just to be clear, you have

r^n + \frac{1}{1 -r} on your right hand side?
 
oopps, sorry. it should be

1+r+...+r^n=1-r^(n+1) / 1-r
 
What is the base case? Let's start with that
 
Also, shouldn't |r|<1?
 
the what case?
 
Base Case. You know that step you do in Induction
 
  • #10
i don't know?
 
  • #11
What can n be and cannot be?
 
  • #12
edit; mistake
 
  • #13
is it n can't be 0? but can be anything else?
 
  • #14
Roscoe1989 said:
oopps, sorry. it should be

1+r+...+r^n=1-r^(n+1) / 1-r

If you don't know induction, just multiply out (1+r+...+r^n)*(1-r). Many terms cancel.
 
  • #15
my textbook is saying 1-r^n+1
 
  • #16
Roscoe1989 said:
my textbook is saying 1-r^n+1

Oh sorry I didn't see the 1 in front earlier
 
Last edited:
  • #17
Roscoe1989 said:
my textbook is saying 1-r^n+1

Your textbook is right.
 
  • #18
Roscoe1989 said:
is it n can't be 0? but can be anything else?

try pluggin in 0
 
  • #19
n has to be greater than 0
 
  • #20
Roscoe1989 said:
n has to be greater than 0

Why?
 
  • #21
1 + r + r^2 + ... + r^n = \frac{1 - r^{n+1}}{1-r}

Is what you have. What happens if n = 0?
 
  • #22
then the right hand side will equal 1. right?
 
  • #23
Roscoe1989 said:
then the right hand side will equal 1. right?

And what about the left hand side? What happens there?
 
  • #24
flyingpig said:
Why?

Why not? I would interpret 1+r+...+r^n for n=0 to be 1. (1-r^(0+1))/(1-r)=(1-r)/(1-r)=1. I'm not sure you are leading this in a helpful direction, flyingpig.
 
  • #25
honestly, i don't know
 
  • #26
Dick said:
Why not? I would interpret 1+r+...+r^n for n=0 to be 1. (1-r^(0+1))/(1-r)=(1-r)/(1-r)=1. I'm not sure you are leading this in a helpful direction, flyingpig.

Nononon, Ros asked if it can be 0 before
 
  • #27
does it mean that it's "proven"?
 
  • #28
Roscoe1989 said:
honestly, i don't know

1 + r + r^2 + ... + r^n = \frac{1 - r^{n+1}}{1-r}

You were originally given this. You got it right when you said the rhs is 1 when n = 0
 
  • #29
Roscoe1989 said:
does it mean that it's "proven"?

By Induction? No

If only though, if only...
 
  • #30
but what does it mean on the LHS?
 

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