Proof Question: Using Mathematical Induction

[tiny-tim]

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 I am 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 … ?

 

[Lococard]

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]

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 = … ?

 

[Lococard]

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

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

 

[tiny-tim]

Yes!

= … ?

 

[tiny-tim]

… 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 …? ​

 

[Schrodinger's Dog]

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 $\neq$ 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. 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.