- #26

tiny-tim

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I don't follow

From the LHS to the RHS of what?

From the LHS to the RHS of what?

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- #26

tiny-tim

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I don't follow

From the LHS to the RHS of what?

From the LHS to the RHS of what?

- #27

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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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[tex]\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}[/tex] = 0

Where did the "= 0" come from?

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 … ?

- #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]

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

is that right?

Yes!

Would i then expand:

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

Noooo …

then after cancelling:

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

Cancelling what?

(And how did you get that?)

The whole plan was to add 1/(n+1)(n+2) to it first …

- #31

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[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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[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?

I liked him!

erm … just read what you've written …

- #33

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

But im back now.

I can tell the difference!

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

And from then i simplfy it?

Yes!

- #35

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[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!

(You

Now im not sure what to do :S

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

It's the same as [itex]\frac{(n+1)}{(n+1)}\,\frac{(n+1)}{(n+2)}[/itex] , which is … ?

- #37

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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!

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

Hint: 1 = … ?

- #39

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

= [(n+2) - 1] / (n+2)?

= [(n+2) - 1] / (n+2)?

- #40

tiny-tim

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Yes!

= … ?

= … ?

- #41

tiny-tim

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

erm …

… êtes-vous sûr …?

- #42

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erm …

… êtes-vous sûr …?

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.

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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