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

- anonymous

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

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- anonymous

Ok so I know that you're supposed to use this equation:
Tsub r+1 = \[\left(\begin{matrix}n \\ r\end{matrix}\right)\]\[a^{n-r}b^{r}\]

- anonymous

I dunno why that came out so weird. Anyway, I plugged in a, so 2x^2, b, and n, but my answer is completely wrong. It's 60 but I got r=6

- anonymous

yes, and your job is to figure out what \(r\) is

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## More answers

- anonymous

yea but I did it wrong. I tried to find the general term but..I don't think my algebra is right

- anonymous

\(n=6\) in your example

- anonymous

yes, n=6 in the example

- anonymous

and what you are looking for is where the exponent on the \(\frac{1}{x}\) term is equal to the exponent on the \(x^2\) term so that you will get 1 when you divide

- anonymous

umm yes, I guess so

- anonymous

sure because you will have a term that looks like \[C\frac{x^m}{x^m}\] which is just \(C\)

- anonymous

I've never seen that equation before but ok

- anonymous

we can do it using algebra, or we can just guess what it has to be

- anonymous

it is not an equation, i am just saying if the exponent is the same top and bottom then the x terms will cancel and we will have a constant

- anonymous

no lets use algebra lol

- anonymous

ok cool

- anonymous

so the first term is in the parentheses is \(2x^2\) meaning you will have \[(2x^2)^{6-r}\] or \[2^{6-r}x^{12-2r}\] for each of those
clear or no?

- anonymous

yup

- anonymous

i know you distribute but do you get 64x^12-2r?

- anonymous

if you write \(\frac{1}{x}\) as \(x^{-1}\) then those terms will look like \(x^{r}\)

- anonymous

no it is not 64 because we don't know what r is

- anonymous

wait I'm confused. so you don't distribute the 6 to the 2?

- anonymous

my book shows that even if you don't know what r is you still distribute it

- anonymous

ok 2 is being raised to the power of \(6-r\)

- anonymous

wait so does it just stay \[2x^{12-2r}\]

- anonymous

nope you have to raise 2 to the power of \(6-r\) as well

- anonymous

right so then it's 64

- anonymous

just like say \[(ab^2)^4=a^4b^8\]

- anonymous

it is not 64 since we do not know what \(r\) is yet

- anonymous

...um could you just show me what it looks like cos I'm confused

- anonymous

i did but i will be happy to write it again \[\huge (2x^2)^{6-r}=2^{6-r}(x^2)^{x-r}=2^{6-r}x^{12-2r}\]

- anonymous

ok makes sense, thank you

- anonymous

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

- anonymous

and the other one will be \[\large x^{-r}\]

- anonymous

when you multiply of course you add the exponents, and you want that product to be 0 because \(x^0=1\) then you will get your constant

- anonymous

wait im confused, why is x^-r?

- anonymous

why is the 6 not added?

- anonymous

or distributed in this case

- anonymous

\[\frac{1}{x}=x^{-1}\] so \[(\frac{1}{x})^r=x^{-r}\]

- anonymous

ohhhh ok thanks

- anonymous

don't forget you wrote \[a^{n-r}b^{r}\] here \(n=6\) so it is \[a^{6-r}b^r\]no 6 in the second part

- anonymous

kk

- anonymous

next job is to solve \[12-2r-r=0\] for \(r\)

- anonymous

what about the 6-r?

- anonymous

you said you wanted to do it the algebra way, so that is what we are doing

- anonymous

also where did the third -r come from in the 12-2r-r equation?

- anonymous

the \(6-r\) got turned in to \(12-2r\) when we distributed the 2 from the \(x^2\) term

- anonymous

that was this line here \[\large (2x^2)^{6-r}=2^{6-r}(x^2)^{x-r}=2^{6-r}x^{12-2r}\]

- anonymous

ok I'm sorry I'm really confused. In the equation you gave me about, you said \[2^{6-r}\]. but then you said you dont distribute the 6 to the 2, so it therefore won't become 64, so you leave it as that. so i don't know how you're getting 6-r into 12-2r if you weren't supposed to do anything that it in the first place

- anonymous

lets go slower

- anonymous

suppose for example \(r=2\)

- anonymous

you would have \[\binom{6}{2}(2x^2)^4(-\frac{1}{x})^2\]

- anonymous

ok um, it's almost 11 where I'm at, and we've been doing this for 20 minutes, so I think I'll just come back later. sorry to have wasted your time but i need to continue. ill be back in a while but im going to close the question because this is just confusing me. thanks again

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