anonymous
  • anonymous
Find the constant term in the expansion of (2x^2 - 1/x)^6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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
anonymous
  • anonymous
@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?
anonymous
  • anonymous
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

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anonymous
  • anonymous
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
anonymous
  • anonymous
pretty much exactly what @peachpi said
anonymous
  • anonymous
since i remember that you are writing this as \[\binom{n}{r}a^{n-r}b^r\]
anonymous
  • anonymous
in the expansion of \[\left( x+a \right)^n\] general term \[T _{r+1}=C_{r}^{n}x ^{n-r}a^r\]
anonymous
  • anonymous
Here \[T _{r+1}=C _{r}^{6}\left( 2 x^2 \right)^{6-r}\left( -\frac{ 1 }{ x } \right)^r\]
anonymous
  • anonymous
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.
1 Attachment
anonymous
  • anonymous
hi ok back
anonymous
  • anonymous
@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
anonymous
  • anonymous
@satellite73 but the answer at the back of my book says "60". I don't understand how they found that
anonymous
  • anonymous
@satellite73 I mean, I did binomial coefficient, so it ended up looking like the following \[\frac{ 6! }{ 4!(6!-4!) } = \]
anonymous
  • anonymous
=15
anonymous
  • anonymous
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
anonymous
  • anonymous
oh btw @peachpi if I do the binomial expansion of this really quick, will you tell me if I got it right?
anonymous
  • anonymous
ahh hello?
zepdrix
  • zepdrix
Figure it out? :)
anonymous
  • anonymous
no ;-; pls help
zepdrix
  • zepdrix
you -_-
anonymous
  • anonymous
we can find the 60
anonymous
  • anonymous
im sorry lol I just don't know what I'm doing. and i really don't want to do pascal's triangle.
anonymous
  • anonymous
ok!
anonymous
  • anonymous
first off we know \(r=4\) right
anonymous
  • anonymous
yes
zepdrix
  • zepdrix
Were you able to set up your general term? :d
zepdrix
  • zepdrix
Err whatever method you were using
anonymous
  • anonymous
yes, it's 12-3r, which is 4
anonymous
  • anonymous
and of course \(n=6\) so on think you need to compute it \(\binom{6}{4}\)
anonymous
  • anonymous
yeah actually i did that on a calculator but i don't want to do the calculator way
anonymous
  • anonymous
\[\binom{6}{4}=\binom{6}{2}=\frac{6\times 5}{2}=15\]
anonymous
  • anonymous
basically, i know that this equals 4, right? so then, like i said, i did the binomial coefficient (look above)
anonymous
  • anonymous
yes, i did that. do you multiply 15 by 4?
anonymous
  • anonymous
because you also have \((2x)^2=4x^2\)
zepdrix
  • zepdrix
\[\large\rm (2x^2-x^{-1})^6\quad=\sum_{k=0}^{6}\left(\begin{matrix}6 \\ \rm k\end{matrix}\right)(2x^2)^{6-k}(-x^{-1})^k\]Are you forgetting about the other numbers that make up your coefficient?
anonymous
  • anonymous
that is where the 4 comes from
zepdrix
  • zepdrix
Ya, namely that fancy 2 there :d
anonymous
  • anonymous
uh...I just got so confused lol
anonymous
  • anonymous
ack not again !!
anonymous
  • anonymous
@zepdrix do you plug in the 4 for the k?
anonymous
  • anonymous
we know \(\binom{6}{4}=15\) right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and the first term is not \(x^2\) it is \(2x^2\) right?
zepdrix
  • zepdrix
\[\large\rm \left(\begin{matrix}6 \\ \rm k\end{matrix}\right)(2x^2)^{6-k}(-x^{-1})^k\]Hopefully the Binomial Theorem stuff I wrote out made sense >.< Ya you could plug in k=4 from this point. Or whatev
anonymous
  • anonymous
you have to square all of \(2x^2\) i.e .\((2x^2)^2=4x^4\)
anonymous
  • anonymous
that is where the 4 comes from
anonymous
  • anonymous
@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?
anonymous
  • anonymous
\[15\times 4=60\]
anonymous
  • anonymous
\[\binom{6}{4}\times 4=60\] in other words
anonymous
  • anonymous
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
anonymous
  • anonymous
thank you so much satellite!!
anonymous
  • anonymous
also thanks zepdrix too lol both of yall are amazing~
anonymous
  • anonymous
actually it is not because you are multiplying by \(r\) it is because you have \[\binom{6}{4}(2x^2)^2\left(\frac{1}{x}\right)^4=15\times 4\]

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