anonymous
  • anonymous
Find the coefficient of x^-6 in the expansion of (2x-3/x^2)^12
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmale
  • mathmale
Are you sure you mean \[x^{-6}?\]
mathmale
  • mathmale
... and not \[x^6?\]
zepdrix
  • zepdrix
Can you clarify the expression you wrote bri, is it this\[\large\rm \left(\frac{2x-3}{x^2}\right)^{12}\] or this\[\large\rm \left(2x-\frac{3}{x^2}\right)^{12}\]

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anonymous
  • anonymous
the latter
anonymous
  • anonymous
ill be back in ten minutes, is that ok? ill just tag you when i come back real quick
zepdrix
  • zepdrix
Ya do whatever you gotta do :d I might post some steps in the mean time.
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
ok I'm back! Yea, I kind of understand that. We have to find the general term, right?
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
kind of..lemme just look at it for a sec to understand it all lol
zepdrix
  • zepdrix
Ya lemme slow down a sec, as a first step I'm applying Binomial Theorem,\[\large\rm (a+b)^n\quad=\sum_{k=0}^n \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k}b^k\]
anonymous
  • anonymous
OHH ok haha I get it. the thing is, I'm in IB and SL math uses r instead of k
zepdrix
  • zepdrix
Ah :)
anonymous
  • anonymous
but ye ok, i get it. also, thanks for separating it all, it makes waaaay more sense that way
zepdrix
  • zepdrix
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,\[\large\rm =\sum_{r=0}^{12}\left(\begin{matrix}12 \\ r\end{matrix}\right)2^{12-r}(-3)^r~~ x^{12-3r}\]
zepdrix
  • zepdrix
Which value of r gives you the x^{-6} term?\[\large\rm (stuff)x^{12-3r}=(stuff)x^{-6}\]
anonymous
  • anonymous
umm..it's the 12-3r right?
anonymous
  • anonymous
cos like thats the general term right..
zepdrix
  • zepdrix
You have to figure out which numerical value of r gives you the x^{-6} on the left side of that equation.
anonymous
  • anonymous
oh ok lol I'm totally lost then
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
ohh ok so then it's 6?
zepdrix
  • zepdrix
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
anonymous
  • anonymous
waait wait wait the back of the book says the answer is something else lemme do the equation version, one sec
anonymous
  • anonymous
\[\left(\begin{matrix}12 \\ 6\end{matrix}\right) * 2^{6}* (-3)^6\]
anonymous
  • anonymous
omg wait
anonymous
  • anonymous
no i get it lol
zepdrix
  • zepdrix
Oh they only wanted the coefficient :) my bad. So we should have left off the x^{-6}.
zepdrix
  • zepdrix
The rest matches up though, yes? :D
anonymous
  • anonymous
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
zepdrix
  • zepdrix
ya i can try c:
anonymous
  • anonymous
thank u haha ill make the new q now

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