Some Binomial Q that I can't do.. (1 Viewer)

study-freak · Apr 27, 2009

Could anyone post the solution for this question plz? Thanks in advance.

study-freak · Apr 27, 2009

Anyone???????????????
(How come my question is being neglected while other maths Q's in this forum are being answered...)

random-1005 · Apr 27, 2009

study-freak said:
Anyone???????????????
(How come my question is being neglected while other maths Q's in this forum are being answered...)

because no1 likes you

Timothy.Siu · Apr 27, 2009

study-freak said:
Anyone???????????????
(How come my question is being neglected while other maths Q's in this forum are being answered...)

question seems dodgey lol

u probably do like
x^2=(1+x)
x^2n=(1+x)^n

but no idea wat after

study-freak · Apr 27, 2009

hmm... this is actually from CSSA Trial Ext 1 very last Q in 2005...

So no one in Boredofstudies can do it?.....
Edit: done by Drongoski

MC Squidge · Apr 27, 2009

dont u have a solution for it with the paper

study-freak · Apr 27, 2009

MC Squidge said:
dont u have a solution for it with the paper

I got it off the internet and it didn't include the solutions.

MC Squidge · Apr 27, 2009

well if u cant do it study freak, no one can

study-freak · Apr 27, 2009

MC Squidge said:
well if u cant do it study freak, no one can

LOL... You've overestimated my maths skills. I'm sure that there are people in Boredofstudies who are much more talented in mathematics than me.

illiterary · Apr 27, 2009

Well, the solutions for x are just the negatives of the values for (x-1). I'm working on it.

Edit: Bah, screw this.

Pwnage101 · Apr 27, 2009

study-freak said:
I got it off the internet and it didn't include the solutions.

could u upload the paper to share wiht the rest of us?

ill try to get to it later

PS. don't get too angry if your question isn't answered in 1 hour, or even 3 hours

give it at least a day.

Drongoski · Apr 27, 2009

Here is my attempt if I'm not too late.

For the benefit of readers, the question was:

The real number x is a solution of the equation: x² - x - 1 = 0. Use the Binomial Theorem to show that the sum
S = 1+x+x²+ . . . + x^2n-1 (n = 1, 2, 3, . . ) is given by: S = sigma (r = 1 to n) of _nC_r times x^r+1.

$\100dpi x^{2}-x-1=0\ \ \Rightarrow \ \ x^{2}= 1+x\ \ \ and\ \ \ \frac{-1}{1-x}=x\\ \\ Now:\ \ S\ =\ 1+x+x^{2}+\ \cdots \ +\ x^{2n-1}\\ \\ \ \ =\ \ \frac{1-x^{2n}}{1-x}\ \ =\ \ \frac{1-(x^{2})^{n}}{1-x}\ \ =\ \ \frac{1-(1+x)^{n}}{1-x}\\ \\ \ \ =\ \ \frac{1\ -\ \sum_{r=0}^{n}\binom{n}{r}x^{r}}{1-x}\ \ =\ \ \frac{-\sum_{r=1}^{n}\binom{n}{r}x^{r}}{1-x}\\ \\ \ \ =\ \ \frac{-1}{1-x}\times \sum_{r=1}^{n}\binom{n}{r}x^{r}\ \ =\ \ x\sum_{r=1}^{n}\binom{n}{r}x^{r}\ \ =\ \ \sum_{r=1}^{n}\binom{n}{r} x^{r+1}$

QED

Drongoski · Apr 27, 2009

study-freak said:
hmm... this is actually from CSSA Trial Ext 1 very last Q in 2005...

So no one in Boredofstudies can do it?.....

You don't accept my solution ?

wixxy2348 · Apr 27, 2009

Drongoski said:
Here is my attempt if I'm not too late.

$\100dpi x^{2}-x-1=0\ \ \Rightarrow \ \ x^{2}= 1+x\ \ \ and\ \ \ \frac{-1}{1-x}=x\\ \\ Now:\ \ S\ =\ 1+x+x^{2}+\ \cdots \ +\ x^{2n-1}\\ \\ \ \ =\ \ \frac{1-x^{2n}}{1-x}\ \ =\ \ \frac{1-(x^{2})^{n}}{1-x}\ \ =\ \ \frac{1-(1+x)^{n}}{1-x}\\ \\ \ \ =\ \ \frac{1\ -\ \sum_{r=0}^{n}\binom{n}{r}x^{r}}{1-x}\ \ =\ \ \frac{-\sum_{r=1}^{n}\binom{n}{r}x^{r}}{1-x}\\ \\ \ \ =\ \ \frac{-1}{1-x}\times \sum_{r=1}^{n}\binom{n}{r}x^{r}\ \ =\ \ x\sum_{r=1}^{n}\binom{n}{r}x^{r}\ \ =\ \ \sum_{r=1}^{n}\binom{n}{r} x^{r+1}$

QED

This, I'm pretty sure.

study-freak · Apr 27, 2009

Drongoski said:
Here is my attempt if I'm not too late.

$\100dpi x^{2}-x-1=0\ \ \Rightarrow \ \ x^{2}= 1+x\ \ \ and\ \ \ \frac{-1}{1-x}=x\\ \\ Now:\ \ S\ =\ 1+x+x^{2}+\ \cdots \ +\ x^{2n-1}\\ \\ \ \ =\ \ \frac{1-x^{2n}}{1-x}\ \ =\ \ \frac{1-(x^{2})^{n}}{1-x}\ \ =\ \ \frac{1-(1+x)^{n}}{1-x}\\ \\ \ \ =\ \ \frac{1\ -\ \sum_{r=0}^{n}\binom{n}{r}x^{r}}{1-x}\ \ =\ \ \frac{-\sum_{r=1}^{n}\binom{n}{r}x^{r}}{1-x}\\ \\ \ \ =\ \ \frac{-1}{1-x}\times \sum_{r=1}^{n}\binom{n}{r}x^{r}\ \ =\ \ x\sum_{r=1}^{n}\binom{n}{r}x^{r}\ \ =\ \ \sum_{r=1}^{n}\binom{n}{r} x^{r+1}$

QED

Thanks a lot

study-freak · Apr 27, 2009

Pwnage101 said:
could u upload the paper to share wiht the rest of us?

If you want

shaon0 · Apr 27, 2009

study-freak said:
If you want

Thanks

Some Binomial Q that I can't do.. (1 Viewer)

study-freak

Bored of

study-freak

Bored of

random-1005

Banned

Timothy.Siu

Prophet 9

study-freak

Bored of

MC Squidge

BOS' Apex Predator

study-freak

Bored of

MC Squidge

BOS' Apex Predator

study-freak

Bored of

illiterary

Member

Pwnage101

Moderator

Drongoski

Well-Known Member

Drongoski

Well-Known Member

wixxy2348

goldmambo #5.

study-freak

Bored of

study-freak

Bored of

shaon0

...

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