Mathematical Induction - Algebra Manipulation

  • Thread starter Modulus85
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  • #1
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Homework Statement


I am working on a mathematical induction problem, where I need to prove:

(1 - (-7) ^ (k + 2)) / 4


Homework Equations



(1 - (-7) ^ (k + 1)) / 4 + 2(-7) ^ (k + 1)

The Attempt at a Solution



So I just need to add the two items in section two above. Now I know I need a common denominator (4):

(1 - (-7) ^ (k + 1)) / 4 + 8(-7) ^ (k + 1) / 4

Unfortunately, I don't know how to progress any further. I don't know how to add the above together. Could someone please provide a few hints to move me along?
 
Last edited:

Answers and Replies

  • #2
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Homework Statement


I am working on a mathematical induction problem, where I need to prove:

(1 - (-7) ^ (k + 2)) / 4


Homework Equations



(1 - (-7) ^ (k + 1)) / 4 + 2(-7) ^ (k + 1)

The Attempt at a Solution



So I just need to add the two items in section two above. Now I know I need a common denominator (4):

(1 - (-7) ^ (k + 1)) / 4 + 8(-7) ^ (k + 1) / 4

Unfortunately, I don't know how to progress any further. I don't know how to add the above together. Could someone please provide a few hints to move me along?
What exactly are you asked to prove?
 
  • #3
Mentallic
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I am working on a mathematical induction problem, where I need to prove:

(1 - (-7) ^ (k + 2)) / 4
You need to prove an expression?
Can you prove x2?
 
  • #4
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I am trying to prove that:
2 - 2 * 7 + 2 * 7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4

n is a non-negative integer

I did the basis step

I'm at the point where I need to prove P(k) -> P(k + 1)
Therefore, swapping in k+1 for n on the right handside above yields:
(1 - (-7) ^ (k + 2)) / 4

Now I need to add 2(-7)^(k+1) to the right side:
(1 - (-7) ^ (k + 1)) / 4 + 2(-7) ^ (k + 1)

The issue is that I don't know how to simplify (1 - (-7) ^ (k + 1)) / 4 + 2(-7) ^ (k + 1)
 
  • #5
SammyS
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I am trying to prove that:
2 - 2 * 7 + 2 * 7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4

n is a non-negative integer

I did the basis step

I'm at the point where I need to prove P(k) -> P(k + 1)
Therefore, swapping in k+1 for n on the right hand side above yields:
(1 - (-7) ^ (k + 2)) / 4

Now I need to add 2(-7)^(k+1) to the right side:
(1 - (-7) ^ (k + 1)) / 4 + 2(-7) ^ (k + 1)

The issue is that I don't know how to simplify (1 - (-7) ^ (k + 1)) / 4 + 2(-7) ^ (k + 1)
Yes, you need to assume P(k) , that is, assume [itex]\displaystyle 2+2(-7)+2(-7)^2+2(-7)^3+\dots+2(-7)^{k}=\frac{1-(-7)^{k+1}}{4}\ .[/itex]

From that assumption, you need to show P(k+1), which is:

[itex]\displaystyle 2+2(-7)+2(-7)^2+2(-7)^3+\dots+2(-7)^{k}+2(-7)^{k+1}=\frac{1-(-7)^{k+2}}{4}\ .[/itex]

Added in Edit:

Oh, I see that you already know this.


For [itex]\displaystyle \frac{1-(-7)^{k+1}}{4}+2(-7)^{k+1}=\frac{1-(-7)^{k+2}}{4}\ ,[/itex]

look at simplifying [itex]\displaystyle \frac{1-(-7)^{k+2}}{4}-\frac{1-(-7)^{k+1}}{4}\ .[/itex]
 
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  • #6
Mentallic
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For [itex]\displaystyle \frac{1-(-7)^{k+1}}{4}+2(-7)^{k+1}=\frac{1-(-7)^{k+2}}{4}\ ,[/itex]

look at simplifying [itex]\displaystyle \frac{1-(-7)^{k+2}}{4}-\frac{1-(-7)^{k+1}}{4}\ .[/itex]
My teacher used to always chuck a tantrum whenever we worked with both sides of the equation we were trying to prove. If your teacher is anything like mine and you can only manipulate one side of the equation, then start with the left.

You have [itex]\displaystyle \frac{1-(-7)^{k+1}}{4}+2(-7)^{k+1}[/itex] and you want to make it equal to [itex]\displaystyle \frac{1-(-7)^{k+2}}{4}[/itex]. Well the first thing you should instantly think of is to make it all one fraction - that is, it should be something / 4.
After that, you know that both numerators must be equal because the denominators are equal, hence you're already that much closer to the answer.
 

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