Question

Find the coefficient of x^-6 in the expansion of (2x-3/x^2)^12

Answer 1

Are you sure you mean \[x^{-6}?\]

Answer 2

... and not \[x^6?\]

Answer 3

Can you clarify the expression you wrote bri,
is it this\[\large\rm \left(\frac{2x-3}{x^2}\right)^{12}\]

or this\[\large\rm \left(2x-\frac{3}{x^2}\right)^{12}\]

Answer 4

the latter

Answer 5

ill be back in ten minutes, is that ok? ill just tag you when i come back real quick

Answer 6

Ya do whatever you gotta do :d
I might post some steps in the mean time.

Answer 7

Binomial Theorem tells us we can write it this way,\[\large\rm (2x-3x^{-2})^{12}\quad=\sum_{k=0}^{12}\left(\begin{matrix}12 \\ k\end{matrix}\right)(2x)^{12-k}(-3x^{-2})^{k}\]We'll have to combine these x's I suppose,
so it will be easier to determine which value of k corresponds to the x^{-6} term.

Answer 8

ok I'm back! Yea, I kind of understand that. We have to find the general term, right?

Answer 9

So if you split up the garbage you get something like this,\[\large\rm =\sum_{k=0}^{12}\left(\begin{matrix}12 \\ k\end{matrix}\right)2^{12-k}(-3)^k~~ x^{12-k}x^{-2k}\]And then we can apply exponent rule to combine the x's, ya?

Answer 10

kind of..lemme just look at it for a sec to understand it all lol

Answer 11

Ya lemme slow down a sec,
as a first step I'm applying Binomial Theorem,\[\large\rm (a+b)^n\quad=\sum_{k=0}^n \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k}b^k\]

Answer 12

OHH ok haha I get it. the thing is, I'm in IB and SL math uses r instead of k

Answer 13

Ah :)

Answer 14

but ye ok, i get it. also, thanks for separating it all, it makes waaaay more sense that way

Answer 15

R's are better? :)\[\large\rm =\sum_{r=0}^{12}\left(\begin{matrix}12 \\ r\end{matrix}\right)2^{12-r}(-3)^r~~ x^{12-r}x^{-2r}\]So combining our x's gives us,\[\large\rm =\sum_{r=0}^{12}\left(\begin{matrix}12 \\ r\end{matrix}\right)2^{12-r}(-3)^r~~ x^{12-3r}\]

Answer 16

Which value of r gives you the x^{-6} term?\[\large\rm (stuff)x^{12-3r}=(stuff)x^{-6}\]

Answer 17

umm..it's the 12-3r right?

Answer 18

cos like thats the general term right..

Answer 19

You have to figure out which numerical value of r gives you the x^{-6} on the left side of that equation.

Answer 20

oh ok lol I'm totally lost then

Answer 21

You're essentially boiling it down to this: \(\large\rm 12-3r=-6\)
and solving for r.

You simply want to know what value of r gives you that -6 for the exponent.

Answer 22

ohh ok so then it's 6?

Answer 23

Mmm good, that sounds right! :)

So your answer will be the general term,\[\large\rm \left(\begin{matrix}12 \\ r\end{matrix}\right)2^{12-r}(-3)^r~~ x^{12-3r}\]evaluated at r=6

Answer 24

waait wait wait the back of the book says the answer is something else
lemme do the equation version, one sec

Answer 25

\[\left(\begin{matrix}12 \\ 6\end{matrix}\right) * 2^{6}* (-3)^6\]

Answer 26

omg wait

Answer 27

no i get it lol

Answer 28

Oh they only wanted the coefficient :) my bad.
So we should have left off the x^{-6}.

Answer 29

The rest matches up though, yes? :D

Answer 30

yeah it makes sense! thank you soo much. would you mind helping me with one more? i could make another question if you want so you could get 2 best respoinses

Answer 31

ya i can try c:

Answer 32

thank u haha ill make the new q now