Question

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

Answer 1

Can someone please help? Honestly have no idea what the hell I'm doing.

Answer 2

i think there is a typo here

Answer 3

nope, there is none

Answer 4

i can do the equation for you if you want

Answer 5

oh wait you were right

Answer 6

imagine

Answer 7

it's supposed to be "Find the coefficient of x^-6 in the expansion of \[(2x - 3/x^2)^{12}\]

Answer 8

@satellite73 please don't leave me now man I really need your help

Answer 9

ooh ok htat makes more sense

Answer 10

ok so how do we go about solving this?

Answer 11

\[(2x-\frac{3}{x^2})^{12}\] right ?

Answer 12

yes

Answer 13

it is going to be a bit of a pain, but we can do it
what class?

Answer 14

IB SL math

Answer 15

and ya, I know, it's gonna be long and annoying ;-;

Answer 16

anyway can we start?

Answer 17

real long and real annoying
first lets say what the answer is

Answer 18

\[43,110,144\]

Answer 19

ok now we can figure out how to do it

Answer 20

the back of the book says the answer is \[\left(\begin{matrix}12 \\ 6\end{matrix}\right)\] \[* 2^6 * (-3)^6\]

Answer 21

what terms will give \(x^{-6}\)
that is a cheating answer

Answer 22

but that is ok, maybe we can just figure out how they got that

Answer 23

yeah, IB is really weird. anyway, idk what terms would find that

Answer 24

when you expand , all the terms will look like \[\binom{12}{k}(2x)^k(-\frac{3}{x^2})^{n-k}\]

Answer 25

yes

Answer 26

not sure if that makes any sense
clear or no?

Answer 27

Wait I don't think that's right? it should be (2x)^n-r

Answer 28

am i wrong?

Answer 29

lol don't be a slave to notation dear, your book uses \(r\) i guess, i used \(k\)
same difference as the kids say

Answer 30

no no no I mean you switched the order

Answer 31

guess your book writes it as \[\binom{12}{r}(2x)^r(-\frac{3}{x^2})^{n-r}\]

Answer 32

no, here I'll show you what they have

Answer 33

oh i see what what you mean no need to write it

Answer 34

ok lol

Answer 35

they must write \[\binom{12}{r}(2x)^{n-r}(-\frac{3}{x^2})^{r}\]

Answer 36

is that correct?

Answer 37

yes

Answer 38

I tried to do the shortcut way and plug in 6 but the negative is what's throwing me off :'(

Answer 39

ok a couple typos there, nvm

Answer 40

mkay

Answer 41

so the real question is, what will give \(x^{-6}\) i.e. what is \(r\) in this case?

Answer 42

6?

Answer 43

or x/6?

Answer 44

you have powers of \(x^2\) in the denominator and powers of \(x\) in the numerator

Answer 45

sorry man I'm lost

Answer 46

ok lets go slow

Answer 47

each term will look something like \[C\frac{(2x)^k}{(-3x^2)^{12-k}}\]

Answer 48

gotcha

Answer 49

like for example \[C\frac{(2x)^7}{(-3x^2)^{5}}\]

Answer 50

mhm

Answer 51

is that clear or no?

Answer 52

ok good

Answer 53

yup!

Answer 54

so the question is, when will the exponent just be \(-6\)

Answer 55

in the example i did above, the exponent in the numerator is \(7\) and in the denominator it is \(10\) so all in all you will have \(x^{-3}\)

Answer 56

ok

Answer 57

now that we see that (i hope) we can go ahead and figure out what \(r\) is

Answer 58

hehe yea i got it

Answer 59

ok so we have to find what would equal -6?

Answer 60

no what i wrote was wrong sorry

Answer 61

ok no biggie

Answer 62

\[\frac{x^r}{x^{2({12-r)}}}=x^{-6}\]

Answer 63

so..we still need to find the numbers or whatever that'll equal -6 right?

Answer 64

exactly

Answer 65

soooo do we pick random numbers or is there something else we can do?

Answer 66

i actually did not use algebra, i used trial and error, because it is pretty easy to see that \(r=-6\) because \[\frac{x^6}{x^{12}}=x^{-6}\]

Answer 67

but we would use algebra if you needed to

Answer 68

I mean I just wanna know what you plugged into r to make that happen. cos I know obviously that 6-12 = -6 but

Answer 69

but I thought that wasn't the formula we were using

Answer 70

yes that is what i plugged in and now that we know \(r=-6\) the answer they wrote should be clear
you want to use algebra?

Answer 71

just lemme look at it for a sec

Answer 72

ok

Answer 73

take your time, i will get a drink

Answer 74

so you plugged in -6 for the r's?

Answer 75

actually ya I wanna see the algebraic way. I know you did the easy way but I don't get how you're doing that

Answer 76

ok sure

Answer 77

first, forgetting the coefficents, (because we are just trying to find \(r\)) we know the terms look like \[\frac{x^r}{(x^2)^{12-r}}\] or
\[\huge \frac{x^r}{x^{24-2r}}\] right?

Answer 78

yessir

Answer 79

if we were a bit more sophisticated we would have subtracted the exponents already (i was only trying to explain so it made sense) and say that each term looks like \[\huge x^{3r-24}\]

Answer 80

yeah I get that

Answer 81

hope you see that i subtracted the exponents as always when you divide, and \[r-(24-2r)=3r-24\]

Answer 82

ya lol

Answer 83

you want that to equal \(-6\) so set \[3r-24=-6\] and solve for \(r\)

Answer 84

you get \(r=6\) pretty much in your head

Answer 85

6

Answer 86

omg wait so we just found the general term

Answer 87

i mean we found r but we needed to find the general term first to do that

Answer 88

right
now that is NOT how i did it, i just wrote it down on paper, and saw that 6 worked to give \(-6\) as the exponents, but that is the algebra that does it

Answer 89

ok so that helped a ton. thanks. but how did they get the answer they got?

Answer 90

i mean i understand everything of the answer up until the (-3)^6 part

Answer 91

ok now we know \(r=6\) right?

Answer 92

yes

Answer 93

so we know the binomial coefficient will be \[\binom{12}{6}\] right ?

Answer 94

yes

Answer 95

and we don't have \(x\) in the numerator, we have \(2x\) so if \(r=6\) and of course that makes \(12-r=6\) also, you have \[\binom{12}{6}2^6(-3)^6x^{-6}\]

Answer 96

umm..gimme a second

Answer 97

just to look at it

Answer 98

if it was \[(x+\frac{1}{x^2})^{12}\]THEN we would have \[\binom{12}{6}x^{-6}\]