#1
Hi guys, I'm struggling to find out how to do this question and I'm wondering if any of you can help?

Prove that:

thanks in advance
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#4
first step...remove the fractions...multiply all by the lowest common denominator thingy....
#5
Just rip out a page of a dictionary and tape it over the paper, and there is your answer.
#6
make all the denominators equal to eachother.

the rest should lead itself to the answer cuz that is a very very simple proof.
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#7
im not rly good with this but you could start with the right side of the equation and turn the denominator 4n/n^2 - 1 into 4n/(n+1)(n-1)
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#8
First of all, 1=(n^2-1)/(n^2-1)

(n+1)/(n-1)=((n+1)(n+1)/(n-1)(n+1))=(n^2+2n+1)/(n^2-1)

2n/(n+1)=((2n(n-1))/((n+1)(n-1))=(2n^2-2n)/(n^2-1)

Then just add.
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#9
you must have a terrible teacher matt
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#10
Multiply top and bottom of first fraction by....

Ach you know what, to hell with it, do your own goddamn work
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#11
Quote by matt92l
Hi guys, I'm struggling to find out how to do this question and I'm wondering if any of you can help?

Prove that:

thanks in advance


I have the exact answer but i have to scan it and email it. when do you need it by
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#12
Quote by tona_107
I have the exact answer but i have to scan it and email it. when do you need it by

well whenever you get a chance if ya can, no mad rush, tonight if it's possible?
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Im a Single Coil man
#13
Quote by matt92l
well whenever you get a chance if ya can, no mad rush, tonight if it's possible?

it actually might not be the same thing but is that a proof for a telescoping sum?
someone posted the solution on here already though.

yeah i wasn't thinking o the same thing.... sorry, i thought that was something else. my bad.
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Last edited by tona_107 at Feb 15, 2009,
#14
still struggling to understand it, can anyone post a really simplified, step by step version of how to do it? thanks so much, i just really don't get proofs
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Im a Single Coil man
#16
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impossible
I got the same answers to my informatics proofs
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#18
hey man is it an exercise in using difference of 2 squares?

with a proof the idea is to manipulate one side until it equals the other.

in this case you have 3 terms on LHS but only one on RHS, so you need to somehow group all of the LHS terms. when fractions are involved the best way is a common denominator.

difference of 2 squares goes:

a^2 - b^2 = (a+b)*(a-b) --> in this case: n^2 - 1 = (n+1)*(n-1)
expand it if you dont believe me

you see you have (n-1) and (n+1) as denominators on LHS and (n^2 - 1) as the denominator on RHS, and this is what you want to get to.

first multiply the top and bottom of the 1st term by (n+1), to give (n-1)*(n+1) as the denominator

then multiply the top and bottom of the second term by (n-1), again to give (n-1)*(n+1) as the denominator.

using the difference of 2 squares you can change the (n+1)*(n-1) into n^2 - 1

the first 2 terms can be easily added as they have the same denominator

this leaves the +1 term on its own, but if you multiply this by (n^2 - 1)/(n^2 - 1), you give it the same denominator as the rest of the terms, allowing you to easily add it to the rest.

before you expand the terms you should get:

(n+1)^2 - 2n(n-1) + (n^2 -1) as the numerator on the LHS, with n^2 -1 as the denominator, and if you expand this out and cancel it down right, you should be left with 4n on the top of the fraction, and the LHS=RHS and then you've proved it.


the lessons to take away from this:
difference of 2 squares
adding fractions using common denominators

hope this makes sense man ,and good luck to you
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#19
^^ What this guy said. I tried to say that a few posts up but I obviously didn't explain it clearly enough.
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#20


alright i've rotated it to save you the trouble now.
Last edited by daryle_goh at Feb 16, 2009,
#21
Brilliant guys, thanks so much!
American Fender Strat
1979 Marshall Superlead
Real Mccoy Wizard Wah
Robert Keeley Boss Tremolo
Euthymia Crucible Fuzz
Fulltone OCD+Deja Vibe
Line 6 DL4
Analogman BC108
Prescription C.O.B

Im a Single Coil man