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anonymous
 one year ago
How to approach this?
anonymous
 one year ago
How to approach this?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Examine if \[(x ^{2}\frac{ 1 }{ \sqrt{x} })^{30}\] contain a constant term and if so, decide it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it will be a number when you have the same exponent in the numerator and denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know I have to use the binomial theorem in some how but still, im looking for a constant term in this? \[(x ^{2}\frac{ 1 }{ \sqrt{x} })^{30}=\left(\begin{matrix}30 \\ 0\end{matrix}\right)*(x ^{2})^{30}+\left(\begin{matrix}30 \\ 1\end{matrix}\right)*(x^2)^{29}+...+\left(\begin{matrix}n \\ r\end{matrix}\right)*(x ^{2}) ^{nr}(\frac{ 1 }{ \sqrt{x} })^{r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have ignored the second term in your expansion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like i said, should be a constant when the power in the numerator is equal to the power in the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0power in the numerator will be \(2(30k)\) and in the denominator will be \(\frac{1}{2}k\) or vice versa

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for what value of \(k\) is \(2(30k)=\frac{1}{2}k\) or ]\(r\) or whatever your variable is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes there is another one as well, but is it clear why that will give a constant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no im not quite following

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\binom{30}{24}\frac{x^{2\times 6}}{x^{\frac{24}{2}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am off by a minus sign maybe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now it is clear that it is a constant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah cause the x:es = 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no because the exponent in the numerator and denominator are both 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh, yeah but that will be \[x ^{1212}=x^{0}=1\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}\] \[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30r}\left(x^{\frac{1}{2}}\right)^r\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now there may be another one as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yes i changed my mind i think there is try \(r=6\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you still get an exponent of 12 top and bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}30 \\ 6\end{matrix}\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, unless i am way off base, your answer is \[2\times \binom{30}{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is 2 x 30 choose 6 because 30 choose 6 and 30 choose 24 is the same?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 , it cant be r=6 also? \[\frac{ x ^{2*(30r)} }{ x ^{\frac{ r }{ 2 }} }=\frac{ x ^{2*(306)} }{ x ^{\frac{ 6 }{ 2 }} }=\frac{ x ^{48} }{ x ^{3} }\] \[x ^{483}=x ^{45}\] `?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but still \[\left(\begin{matrix}30 \\ 24\end{matrix}\right) \leftarrow \rightarrow \left(\begin{matrix}30 \\ 6\end{matrix}\right)\] right?
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