Author Topic: Phresh's Methods Questions (Read 2190 times) Tweet Share

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

the.watchman · « **Reply #1 on:** February 14, 2010, 03:50:15 pm »

Required term $= ^9C_2 \times (-\frac{x}{4})^2 \times (3)^7 = \frac{19683}{4}x^2$

Phresh · « **Reply #2 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 ?

the.watchman · « **Reply #3 on:** February 14, 2010, 03:52:35 pm »

OOPS, edited post!
Uber-sorry!!!

Phresh · « **Reply #4 on:** February 14, 2010, 03:59:08 pm »

Ahhh thanks, i did a small mistake and screwed up. Thanks for the help

the.watchman · « **Reply #5 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!

Phresh · « **Reply #6 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

the.watchman · « **Reply #7 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...

Phresh · « **Reply #8 on:** February 14, 2010, 05:12:08 pm »

Hrmm yeah that's a way to do it

, Anymore ways ?

the.watchman · « **Reply #9 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 , 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

TrueTears · « **Reply #10 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

the.watchman · « **Reply #11 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!

Phresh · « **Reply #12 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

m@tty · « **Reply #13 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?

superflya · « **Reply #14 on:** February 15, 2010, 11:18:20 pm »

im confused :s

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