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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Phresh on February 14, 2010, 03:48:08 pm

Title: Phresh's Methods Questions
Post by: Phresh on February 14, 2010, 03:48:08 pm: Hello.

Find the third term in the expansion of equation, $(3-x/4)^9$ assuming ascending powers of x.

I got -81/4096x^7 but the answer says 19683x^2/4

TY
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 14, 2010, 03:50:15 pm: Required term $= ^9C_2 \times (-\frac{x}{4})^2 \times (3)^7 = \frac{19683}{4}x^2$
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 14, 2010, 03:52:14 pm: Quote from: the.watchman on February 14, 2010, 03:50:15 pm
Required term $= ^9C_3 \times (-\frac{x}{4})^3 \times (3)^6 = -\frac{15309}{16}x^3$
Isn't the question asking for the third term? not the coefficient of x^3 ?
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 14, 2010, 03:52:35 pm: OOPS, edited post!
Uber-sorry!!! :P
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 14, 2010, 03:59:08 pm: Ahhh thanks, i did a small mistake and screwed up. Thanks for the help :)
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 14, 2010, 04:00:11 pm: Quote from: Phresh on February 14, 2010, 03:59:08 pm
Ahhh thanks, i did a small mistake and screwed up. Thanks for the help :)

No prob, good luck! :)
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 14, 2010, 04:22:20 pm: Another question

Find the coefficient of y^4 in the expansion of $(y+3)^3(2-y)^5$

What is the easiest way to approach these questions?

Thanks
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 14, 2010, 04:28:58 pm: Split it into two, then maybe find 0,4 and 1,3 and 2,2 and 3,1

So for binomial 1:

Co-efficient of $y^0$ $= 3^3 = 27$

Co-efficient of $y^1$ $= ^3C_1 \times (3)^2 = 27$

Co-efficient of $y^2$ $= ^3C_2 \times (3)^1 = 9$

Co-efficient of $y^3$ $= ^3C_3 \times (3)^0 = 1$

Then for binomial 2:

Co-efficient of $y^1$ $= ^5C_1 \times (-1) \times (2)^4 = -80$

Co-efficient of $y^2$ $= ^5C_2 \times (2)^3 = 80$

Co-efficient of $y^3$ $= ^5C_3 \times (-1) \times (2)^2 = -40$

Co-efficient of $y^4$ $= ^5C_4 \times (2)^1 = 10$

Now we can find the overall co-efficient:

Co-efficient $= (27)(10) + (27)(-40) + (9)(80) + (1)(-80) = -170$

SO the answer is -170, but there's gotta be an easier way... :P
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 14, 2010, 05:12:08 pm: Hrmm yeah that's a way to do it :D, Anymore ways ?
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 14, 2010, 05:13:46 pm: Quote from: Phresh on February 14, 2010, 05:12:08 pm
Hrmm yeah that's a way to do it :D, Anymore ways ?

Not that I know of, but I could be wrong.
If there was a way to factorise it, then you could shortcut, I guess.
But that doesn't apply to this one
Title: Re: Phresh's Methods Questions
Post by: TrueTears on February 14, 2010, 05:18:40 pm: Phresh have a read here: (I remember StudyingHard asked the same question.)

http://vcenotes.com/forum/index.php/topic,21221.msg215736.html#msg215736

You should also have a read of the multinomial theorem.

http://vcenotes.com/forum/index.php/topic,19896.msg205887.html#msg205887
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 14, 2010, 05:21:28 pm: Quote from: TrueTears on February 14, 2010, 05:18:40 pm
Phresh have a read here:

http://vcenotes.com/forum/index.php/topic,21221.msg215736.html#msg215736

Genius! :D
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 15, 2010, 10:20:47 pm: -3(x+11/6)^2 -161/36

do we have to incorporate -3 back into the brackets etc.. ?

Ty
Title: Re: Phresh's Methods Questions
Post by: m@tty on February 15, 2010, 11:14:36 pm: Quote from: Phresh on February 15, 2010, 10:20:47 pm
-3(x+11/6)^2 -161/36

do we have to incorporate -3 back into the brackets etc.. ?

Ty

$\begin{align} -3 \left(x+ \frac{11}{6} \right)^2- \frac{161}{36} &= -\left( \sqrt{3}\left(x+ \frac{11}{6}\right)\right)^2- \frac{161}{36} \\ &= -\left( \sqrt{3}x+ \frac{11\sqrt{3}}{6}\right)^2- \frac{161}{36}\end{align}$

I don't know why you'd want to do that though. Why do you need to do this?
Title: Re: Phresh's Methods Questions
Post by: superflya on February 15, 2010, 11:18:20 pm: im confused :s
Title: Re: Phresh's Methods Questions
Post by: TrueTears on February 15, 2010, 11:25:44 pm: r u trying to apply DOPS?

u r better off factoring out the 3 then making it into a square
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 16, 2010, 05:37:47 am: Quote from: Phresh on February 15, 2010, 10:20:47 pm
-3(x+11/6)^2 -161/36

do we have to incorporate -3 back into the brackets etc.. ?

Ty

For solving or anything, just leaving it like that is probably what you want, putting the -3 back in seems a bit pointless :)
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 18, 2010, 08:45:28 pm: Cheers

in the expansion of (2a-1)^n, the coefficient of the second term is -192. Find the value of n.
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 18, 2010, 08:50:55 pm: So $^nC_1 \cdot (2)^{1} \cdot (-1)^{n-1} = -192$ OR $^nC_1 \cdot (2)^{n-1} \cdot (-1)^{1} = -192$

Because $^nC_1 \cdot (2)^{1} \cdot (-1)^{n-1} = \frac{n!}{1!(n-1)!} \cdot 2 \cdot (-1)^{n-1}$

$^nC_1 \cdot (2)^{1} \cdot (-1)^{n-1} = n \cdot 2 \cdot (-1)^{n-1} = 2n \cdot (-1)^n$

And n>0

Therefore $n=96$
Title: Re: Phresh's Methods Questions
Post by: Phresh on February 18, 2010, 09:18:52 pm: The answer says 6 though :S:S hrmm
Title: Re: Phresh's Methods Questions
Post by: the.watchman on February 18, 2010, 09:22:22 pm: Ah, n=96 does work for the first case that I put forward, however, the second case gives n=6 :D
Why did I get the most complicated answer...