BINOMIAL THEOREM question: http://prntscr.com/8ib6g8 Find the numerically greatest coefficient in the expansion of (2 - x^2/4)^10.

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BINOMIAL THEOREM question: http://prntscr.com/8ib6g8 Find the numerically greatest coefficient in the expansion of (2 - x^2/4)^10.

Mathematics
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I'm confused on what to do with the negative sign?? I know we use the formula: \[\frac{ T _{k+1} }{ T_{k}} = \frac{ n - k + 1 }{ k } \times \frac{ b }{ a }\]
Or do we ignore the negative? O_O
2^10

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I don't know what this is, so I don't know what they mean. I think the binomial is \[(a+b)^n = \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}\] So I'm not so sure what they're wanting here, I am guessing coefficient on x after it's all expanded out right?
This is what i did but i'm not sure if i'm solving this correctly :S http://prntscr.com/8ib6g8
it would be -1280
if negatives dont matter then that is right i think
I know this doesn't really help with the problem all that much, but I would factor the 1/4 out of each term, that is just getting in the way in my opinion,\[\large\rm \left(2-\frac{x^2}{4}\right)^{10}=\frac{1}{4^{10}}\left(8-x^2\right)^{10}\]
Yah I would imagine that's what they meant by "numerically greatest", like ignore the sign :O that sounds right.
sheesh why cant they just say magnitude or abs value -.-
hmm i'm lost xD I don't know whether it will be -1280 or 1280... I think maybe it'll be -1280??
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why x=1? :o does that just make the rest of the calculations easier or something?
i actually have no idea LOLL. I just got taught to set x = 1 if there was no given value of x? So unless the question gave a specific value of x, you'd just plug in x = 1 :/ If they asked for the term independent of x, then you set x = 0
I don't know that weird T formula 0_o I'mma do it the long way and see if that checks out.\[\large\rm \frac{T_{k+1}}{T_k}=\frac{\color{orangered}{\left(\begin{matrix}10 \\ k+1\end{matrix}\right)}2^{10-(k+1)}\left(-\frac{x}{4}\right)^{k+1}}{\color{orangered}{\left(\begin{matrix}10 \\ k\end{matrix}\right)}2^{10-k}\left(-\frac{x}{4}\right)^{k}}\]Which I guess simplifies down a bit right? Ummm\[\large\rm \frac{T_{k+1}}{T_k}=\frac{\color{orangered}{\frac{10!}{(k+1)!(10-(k+1))!}}\left(-\frac{x}{4}\right)^1}{\color{orangered}{\frac{10!}{k!(10-k)!}}2^1}\]And more simplifying -_-\[\large\rm \frac{T_{k+1}}{T_k}=\frac{k!(10-k)!}{(k+1)!(10-(k+1))!}\cdot\left(-\frac{x}{8}\right)\]and further -_-\[\large\rm \frac{T_{k+1}}{T_k}=\frac{(10-k)}{(k+1)}\cdot\left(-\frac{x}{8}\right)\]Hmmmm yah I'm seeing how you got 11 :O Thinking...
1280 i read some post where numerical greatest means the magnitude of the number
Not seeing how you got 11* blah
pls show me appreciation with your medals
is there even a difference between "greatest coefficient" and "numerically greatest coefficient" ? o.o But alrighties, i guess i'll stick with 1280...
Oh your formula works actually,\[\large\rm \frac{T_{k+1}}{T_k}=\frac{(10-\color{royalblue}{(k)})}{(\color{royalblue}{k}+1)}\cdot\left(-\frac{x}{8}\right)\] \[\large\rm \frac{T_{k+1}}{T_k}=\frac{(10-\color{royalblue}{(k-1)})}{(\color{royalblue}{k-1}+1)}\cdot\left(-\frac{x}{8}\right)\] \[\large\rm \frac{T_{k+1}}{T_k}=\frac{10-k+1}{k}\cdot\left(-\frac{x}{8}\right)\]Ok I'll simmer down XD
yes greatest coefficient means the biggest number where positive numbers are greater than the negative numbers Numerically greatest means the number that is farthest away from 0 or the origin
LMAO Zepdrix xD Your dedication is so strong. Binomials give me headaches. Dan, wait, so if it is the number farthest away from 0 or the origin, wouldn't it be -1280?
yes thats right
okaaayyy ^_^ Thank you ALL~ <3.
no problem
noob
-.-
ooo snap
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*

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