Proof Question: Using Mathematical Induction

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  • #26
tiny-tim
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I don't follow :confused:

From the LHS to the RHS of what?
 
  • #27
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[tex]\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex] = 0

n/(n+1) = n(n+2)/(n+1)(n+2).


Did you swap [tex]\frac{1}{(n+1)(n+2)}[/tex] to the other side of the equals sign?

How did you simplify did you get n(n+2)/(n+1)(n+2). Is it all on the LHS of the equals sign, or did you swap some of it to the right side?
 
  • #28
tiny-tim
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Sorry, you've completely lost me:
[tex]\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex] = 0

Where did the "= 0" come from? :confused:

There's no 0 in this.
Did you swap [tex]\frac{1}{(n+1)(n+2)}[/tex] to the other side of the equals sign?

How did you simplify did you get n(n+2)/(n+1)(n+2). Is it all on the LHS of the equals sign, or did you swap some of it to the right side?

I just did: [tex]\frac{n}{(n+1)}\,=\,\frac{n}{(n+1)}\,\frac{(n+2)}{(n+2)}\,=\,\frac{n(n+2)}{(n+1)(n+2)}\,.[/tex]

That was so that I could then add it to the [tex]\frac{1}{(n+1)(n+2)}[/tex] , giving … ? :smile:
 
  • #29
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[itex]\frac{n(n+2)}{(n+1)(n+2)}[/itex]


is that right?

Would i then expand:

[itex]\frac{n^2+2n}{n^2+3n+2}[/itex]

then after cancelling:

[itex]\frac{1}{(n+2)}[/itex]
 
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  • #30
tiny-tim
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… stick to the plan …

[itex]\frac{n(n+2)}{(n+1)(n+2)}[/itex]

is that right?

Yes! :smile:
Would i then expand:

[itex]\frac{n^2+2n}{n^2+3n+2}[/itex]

Noooo … never expand the bottom !!!! :rolleyes:
then after cancelling:

[itex]\frac{1}{(n+2)}[/itex]

Cancelling what? :confused:

(And how did you get that?)

The whole plan was to add 1/(n+1)(n+2) to it first … then you can try cancelling. :smile:
 
  • #31
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So, what i need to do is:

[itex]\frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1(n+2)}[/itex]

Doesnt that just equal:

[itex]\frac{n(n+2)}{(n+1)(n+2)}[/itex]?
 
  • #32
tiny-tim
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… attack of the clones …

So, what i need to do is:

[itex]\frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1(n+2)}[/itex]

Doesnt that just equal:

[itex]\frac{n(n+2)}{(n+1)(n+2)}[/itex]?

Hey … what happened to the original Lococard? :confused:

I liked him!

:redface: erm … just read what you've written … :redface:
 
  • #33
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haha, Sorry i lost my password and this connection doesnt allow gmail.com to load.

But im back now.

[itex]\frac{n(n+2)+1}{(n+1)(n+2)}[/itex]

And from then i simplfy it?
 
  • #34
tiny-tim
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… long live lococard … !

haha, Sorry i lost my password and this connection doesnt allow gmail.com to load.

But im back now.

I can tell the difference! :biggrin:
[itex]\frac{n(n+2)+1}{(n+1)(n+2)}[/itex]

And from then i simplfy it?

Yes! :smile:
 
  • #35
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So that is the same as:

[itex]\frac{(n+1)(n+1)}{(n+1)(n+2)}?[/itex]

yes? no?

Now im not sure what to do :S
 
  • #36
tiny-tim
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So that is the same as:

[itex]\frac{(n+1)(n+1)}{(n+1)(n+2)}?[/itex]

yes? no?

Yes! :smile:

(You knew that, didn't you?)
Now im not sure what to do :S

You really don't dig polynomial fractions, do you? :frown:

It's the same as [itex]\frac{(n+1)}{(n+1)}\,\frac{(n+1)}{(n+2)}[/itex] , which is … ? :smile:
 
  • #37
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Im really not sure.

You can cancel the (n+1) / (n+1)


Leaving (n+1) / (n+2)?



Im a little confused as the teacher suggested that i work my way to get 1 - [1 / (n+2)]
 
  • #38
tiny-tim
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You can cancel the (n+1) / (n+1)

Leaving (n+1) / (n+2)?

Yes! :smile:
Im a little confused as the teacher suggested that i work my way to get 1 - [1 / (n+2)]

Well, maybe teacher is right …

What is 1 - [1 / (n+2)] ?

Hint: 1 = … ? :smile:
 
  • #39
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(n+2) / (n+2) - [1/(n+2)]?

= [(n+2) - 1] / (n+2)?
 
  • #40
tiny-tim
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Yes!

= … ? :smile:
 
  • #41
tiny-tim
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… merveilleux … !

[tex]= \frac{(n+2) - 1} {(n+2)}? = 1-\frac{1}{n+1} \equiv \sum_{k=1}^n \frac{1}{k(k+1)}[/tex]​

erm …
:redface: … êtes-vous sûr …? :redface:
 
  • #42
erm …
:redface: … êtes-vous sûr …? :redface:

Which of those equals the last one, I left the question marks in there for a good reason, writing which of these is correct in latex gets messy, Ie ?=. The point is that one of those = a third, or the summation given at the beginning of the thread with a telescoping series.

Ie you have your answer. Et voila...

Since obviously some people found this confusing I've edited it to clear that up.

I kept putting [itex]\neq[/itex] in then removing it then putting it in again before I decided it looks better with the question marks. :smile:

If you don't believe me ask a mentor if I edited it about 219283927^34839 times. :tongue:


EDIT: Actually sod it I might as well delete it, as it only makes sense, if you take on board the post about 1/k(k+1) being equivalent to the LHS. So nm.

I was trying to helpfully point out that what gib7 said was the same as what you eventually end up with, but it came out completely wrong.
 
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