#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

Prove that:

thanks in advance

#2

7

duh

duh

#3

Proofs suck. Theres the answer!

#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.

the rest should lead itself to the answer cuz that is a very very simple proof.

#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)

#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.

(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.

#9

you must have a terrible teacher matt

#10

Multiply top and bottom of first fraction by....

Ach you know what, to hell with it, do your own goddamn work

Ach you know what, to hell with it, do your own goddamn work

#11

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

#12

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?

#13

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.

*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

#15

Pi, or 7

#16

Pi, or 7

impossible

I got the same answers to my informatics proofs

#17

impossible

I got the same answers to my informatics proofs

#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

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

#19

^^ What this guy said. I tried to say that a few posts up but I obviously didn't explain it clearly enough.

#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!