anonymous
  • anonymous
How to approach this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Examine if \[(x ^{2}-\frac{ 1 }{ \sqrt{x} })^{30}\] contain a constant term and if so, decide it.
anonymous
  • anonymous
it will be a number when you have the same exponent in the numerator and denominator
anonymous
  • anonymous
I know I have to use the binomial theorem in some how but still, im looking for a constant term in this? \[(x ^{2}-\frac{ 1 }{ \sqrt{x} })^{30}=\left(\begin{matrix}30 \\ 0\end{matrix}\right)*(x ^{2})^{30}+\left(\begin{matrix}30 \\ 1\end{matrix}\right)*(x^2)^{29}+...+\left(\begin{matrix}n \\ r\end{matrix}\right)*(x ^{2}) ^{n-r}(\frac{ 1 }{ \sqrt{x} })^{r}\]

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anonymous
  • anonymous
you have ignored the second term in your expansion
anonymous
  • anonymous
like i said, should be a constant when the power in the numerator is equal to the power in the denominator
anonymous
  • anonymous
power in the numerator will be \(2(30-k)\) and in the denominator will be \(\frac{1}{2}k\) or vice versa
anonymous
  • anonymous
\[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}\]
anonymous
  • anonymous
yeah exactly
anonymous
  • anonymous
so for what value of \(k\) is \(2(30-k)=\frac{1}{2}k\) or ]\(r\) or whatever your variable is?
anonymous
  • anonymous
24?
anonymous
  • anonymous
yes there is another one as well, but is it clear why that will give a constant?
anonymous
  • anonymous
no im not quite following
anonymous
  • anonymous
\[\binom{30}{24}\frac{x^{2\times 6}}{x^{\frac{24}{2}}}\]
anonymous
  • anonymous
i am off by a minus sign maybe
anonymous
  • anonymous
oh no really
anonymous
  • anonymous
now it is clear that it is a constant?
anonymous
  • anonymous
yeah cause the x:es = 1?
anonymous
  • anonymous
no because the exponent in the numerator and denominator are both 12
anonymous
  • anonymous
ooh, yeah but that will be \[x ^{12-12}=x^{0}=1\] ?
anonymous
  • anonymous
\[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}\] \[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}\left(x^{-\frac{1}{2}}\right)^r\]
anonymous
  • anonymous
right
anonymous
  • anonymous
now there may be another one as well
anonymous
  • anonymous
oh no maybe not
anonymous
  • anonymous
oh yes i changed my mind i think there is try \(r=6\)
anonymous
  • anonymous
you still get an exponent of 12 top and bottom
anonymous
  • anonymous
\[\left(\begin{matrix}30 \\ 6\end{matrix}\right)\]
anonymous
  • anonymous
yeah exactly
anonymous
  • anonymous
so, unless i am way off base, your answer is \[2\times \binom{30}{6}\]
anonymous
  • anonymous
It is 2 x 30 choose 6 because 30 choose 6 and 30 choose 24 is the same?
anonymous
  • anonymous
2 * 593775
anonymous
  • anonymous
@satellite73 , it cant be r=6 also? \[\frac{ x ^{2*(30-r)} }{ x ^{\frac{ r }{ 2 }} }=\frac{ x ^{2*(30-6)} }{ x ^{\frac{ 6 }{ 2 }} }=\frac{ x ^{48} }{ x ^{3} }\] \[x ^{48-3}=x ^{45}\] `?
anonymous
  • anonymous
but still \[\left(\begin{matrix}30 \\ 24\end{matrix}\right) \leftarrow \rightarrow \left(\begin{matrix}30 \\ 6\end{matrix}\right)\] right?

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