Question

Find the constant term in the expansion of (2x^2 - 1/x)^6

Answer 1

I'm so lost on this one, I keep seeing people talking about pascal's triangle or doing the binomial coefficient but i honestly don't understand

Answer 2

@Astrophysics Hiii are you busy right now? I don't wanna bother you or anything but do you think you could help me a little with this one?

Answer 3

Since the exponent is 6, you need to use the 7th row of Pascal's triangle:
1, 6, 15, 20, 15, 6, 1

Then you mulitply those coefficients by the two "inside" terms, one in increasing order from 0 to 6, and the other in decreasing order from 6 to 0.

For example, the first term will be \[1(2x)^6\left( -\frac{ 1 }{ x } \right)^0\]
The second will be \[6(2x)^5\left( -\frac{ 1 }{ x } \right)^1\]
The third
\[15(2x)^4\left( -\frac{ 1 }{ x } \right)^2\]
and so on up to the 7th term.
The constant term will be the one without an x

Answer 4

this time we do it the easy way with no algebra
if \(r=4\) and \(n-r=6-4=2\) the exponents will be the same top and bottom
that will give you the constant term

Answer 5

pretty much exactly what @peachpi said

Answer 6

since i remember that you are writing this as \[\binom{n}{r}a^{n-r}b^r\]

Answer 7

in the expansion of \[\left( x+a \right)^n\]
general term
\[T _{r+1}=C_{r}^{n}x ^{n-r}a^r\]

Answer 8

Here
\[T _{r+1}=C _{r}^{6}\left( 2 x^2 \right)^{6-r}\left( -\frac{ 1 }{ x } \right)^r\]

Answer 9

Well, I "cheated"
60
Mathematica can expand the term to the 6th power in 78 micro seconds on
an Apple IMac.
Refer to the attachment.

Answer 10

hi ok back

Answer 11

@robtobey thanks, but I kinda wanted to understand it instead :p

@satellite73 hey again! so nice to see you ^-^ I understand that r=4 after finding the general term. And I know you have to subtract 6 by 4, which gets you 2

Answer 12

@satellite73 but the answer at the back of my book says "60". I don't understand how they found that

Answer 13

@satellite73 I mean, I did binomial coefficient, so it ended up looking like the following

\[\frac{ 6! }{ 4!(6!-4!) } = \]

Answer 14

=15

Answer 15

And so I'm confused if you're supposed to multiply that by 4, which equals 60, which is the answer OR if I'm getting the correct answer only be chance

Answer 16

oh btw @peachpi if I do the binomial expansion of this really quick, will you tell me if I got it right?

Answer 17

ahh hello?

Answer 18

Figure it out? :)

Answer 19

no ;-; pls help

Answer 20

you -_-

Answer 21

we can find the 60

Answer 22

im sorry lol I just don't know what I'm doing. and i really don't want to do pascal's triangle.

Answer 23

ok!

Answer 24

first off we know \(r=4\) right

Answer 25

yes

Answer 26

Were you able to set up your general term? :d

Answer 27

Err whatever method you were using

Answer 28

yes, it's 12-3r, which is 4

Answer 29

and of course \(n=6\) so on think you need to compute it \(\binom{6}{4}\)

Answer 30

yeah actually i did that on a calculator but i don't want to do the calculator way

Answer 31

\[\binom{6}{4}=\binom{6}{2}=\frac{6\times 5}{2}=15\]

Answer 32

basically, i know that this equals 4, right? so then, like i said, i did the binomial coefficient (look above)

Answer 33

yes, i did that. do you multiply 15 by 4?

Answer 34

because you also have \((2x)^2=4x^2\)

Answer 35

\[\large\rm (2x^2-x^{-1})^6\quad=\sum_{k=0}^{6}\left(\begin{matrix}6 \\ \rm k\end{matrix}\right)(2x^2)^{6-k}(-x^{-1})^k\]Are you forgetting about the other numbers that make up your coefficient?

Answer 36

that is where the 4 comes from

Answer 37

Ya, namely that fancy 2 there :d

Answer 38

uh...I just got so confused lol

Answer 39

ack not again !!

Answer 40

@zepdrix do you plug in the 4 for the k?

Answer 41

we know \(\binom{6}{4}=15\) right?

Answer 42

yes

Answer 43

and the first term is not \(x^2\) it is \(2x^2\) right?

Answer 44

\[\large\rm \left(\begin{matrix}6 \\ \rm k\end{matrix}\right)(2x^2)^{6-k}(-x^{-1})^k\]Hopefully the Binomial Theorem stuff I wrote out made sense >.<
Ya you could plug in k=4 from this point.
Or whatev

Answer 45

you have to square all of \(2x^2\) i.e .\((2x^2)^2=4x^4\)

Answer 46

that is where the 4 comes from

Answer 47

@satellite73 yes, i understand where 4 is coming from, i just don't understand where 60 is coming from

@zepdrix ok so like, after that, would I get 60?

Answer 48

\[15\times 4=60\]

Answer 49

\[\binom{6}{4}\times 4=60\] in other words

Answer 50

ok, so yeah, i totally get that then, that was my question before. once you get 15, you multiply it by your r to get 60

Answer 51

thank you so much satellite!!

Answer 52

also thanks zepdrix too lol both of yall are amazing~

Answer 53

actually it is not because you are multiplying by \(r\) it is because you have \[\binom{6}{4}(2x^2)^2\left(\frac{1}{x}\right)^4=15\times 4\]