# Proof Question: Using Mathematical Induction

• Mathematica
tiny-tim
Homework Helper
I don't follow

From the LHS to the RHS of what?

$$\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = 0

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

Did you swap $$\frac{1}{(n+1)(n+2)}$$ 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?

tiny-tim
Homework Helper
Sorry, you've completely lost me:
$$\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}$$ = 0

Where did the "= 0" come from?

There's no 0 in this.
Did you swap $$\frac{1}{(n+1)(n+2)}$$ 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: $$\frac{n}{(n+1)}\,=\,\frac{n}{(n+1)}\,\frac{(n+2)}{(n+2)}\,=\,\frac{n(n+2)}{(n+1)(n+2)}\,.$$

That was so that I could then add it to the $$\frac{1}{(n+1)(n+2)}$$ , giving … ?

$\frac{n(n+2)}{(n+1)(n+2)}$

is that right?

Would i then expand:

$\frac{n^2+2n}{n^2+3n+2}$

then after cancelling:

$\frac{1}{(n+2)}$

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tiny-tim
Homework Helper
… stick to the plan …

$\frac{n(n+2)}{(n+1)(n+2)}$

is that right?

Yes!
Would i then expand:

$\frac{n^2+2n}{n^2+3n+2}$

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

$\frac{1}{(n+2)}$

Cancelling what?

(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.

So, what i need to do is:

$\frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1(n+2)}$

Doesnt that just equal:

$\frac{n(n+2)}{(n+1)(n+2)}$?

tiny-tim
Homework Helper
… attack of the clones …

So, what i need to do is:

$\frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1(n+2)}$

Doesnt that just equal:

$\frac{n(n+2)}{(n+1)(n+2)}$?

Hey … what happened to the original Lococard?

I liked him!

erm … just read what you've written …

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

But im back now.

$\frac{n(n+2)+1}{(n+1)(n+2)}$

And from then i simplfy it?

tiny-tim
Homework Helper
… 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!
$\frac{n(n+2)+1}{(n+1)(n+2)}$

And from then i simplfy it?

Yes!

So that is the same as:

$\frac{(n+1)(n+1)}{(n+1)(n+2)}?$

yes? no?

Now im not sure what to do :S

tiny-tim
Homework Helper
So that is the same as:

$\frac{(n+1)(n+1)}{(n+1)(n+2)}?$

yes? no?

Yes!

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

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

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

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)]

tiny-tim
Homework Helper
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 is 1 - [1 / (n+2)] ?

Hint: 1 = … ?

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

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

tiny-tim
Homework Helper
Yes!

= … ?

tiny-tim
Homework Helper
… merveilleux … !

$$= \frac{(n+2) - 1} {(n+2)}? = 1-\frac{1}{n+1} \equiv \sum_{k=1}^n \frac{1}{k(k+1)}$$​

erm …
… êtes-vous sûr …?

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.

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

I kept putting $\neq$ 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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