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anonymous

  • one year ago

How to approach this?

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  1. anonymous
    • one year ago
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    Examine if \[(x ^{2}-\frac{ 1 }{ \sqrt{x} })^{30}\] contain a constant term and if so, decide it.

  2. anonymous
    • one year ago
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    it will be a number when you have the same exponent in the numerator and denominator

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

  4. anonymous
    • one year ago
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    you have ignored the second term in your expansion

  5. anonymous
    • one year ago
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    like i said, should be a constant when the power in the numerator is equal to the power in the denominator

  6. anonymous
    • one year ago
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    power in the numerator will be \(2(30-k)\) and in the denominator will be \(\frac{1}{2}k\) or vice versa

  7. anonymous
    • one year ago
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    \[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}\]

  8. anonymous
    • one year ago
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    yeah exactly

  9. anonymous
    • one year ago
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    so for what value of \(k\) is \(2(30-k)=\frac{1}{2}k\) or ]\(r\) or whatever your variable is?

  10. anonymous
    • one year ago
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    24?

  11. anonymous
    • one year ago
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    yes there is another one as well, but is it clear why that will give a constant?

  12. anonymous
    • one year ago
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    no im not quite following

  13. anonymous
    • one year ago
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    \[\binom{30}{24}\frac{x^{2\times 6}}{x^{\frac{24}{2}}}\]

  14. anonymous
    • one year ago
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    i am off by a minus sign maybe

  15. anonymous
    • one year ago
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    oh no really

  16. anonymous
    • one year ago
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    now it is clear that it is a constant?

  17. anonymous
    • one year ago
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    yeah cause the x:es = 1?

  18. anonymous
    • one year ago
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    no because the exponent in the numerator and denominator are both 12

  19. anonymous
    • one year ago
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    ooh, yeah but that will be \[x ^{12-12}=x^{0}=1\] ?

  20. anonymous
    • one year ago
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    \[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}\] \[=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}\left(x^{-\frac{1}{2}}\right)^r\]

  21. anonymous
    • one year ago
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    right

  22. anonymous
    • one year ago
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    now there may be another one as well

  23. anonymous
    • one year ago
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    oh no maybe not

  24. anonymous
    • one year ago
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    oh yes i changed my mind i think there is try \(r=6\)

  25. anonymous
    • one year ago
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    you still get an exponent of 12 top and bottom

  26. anonymous
    • one year ago
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    \[\left(\begin{matrix}30 \\ 6\end{matrix}\right)\]

  27. anonymous
    • one year ago
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    yeah exactly

  28. anonymous
    • one year ago
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    so, unless i am way off base, your answer is \[2\times \binom{30}{6}\]

  29. anonymous
    • one year ago
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    It is 2 x 30 choose 6 because 30 choose 6 and 30 choose 24 is the same?

  30. anonymous
    • one year ago
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    2 * 593775

  31. anonymous
    • one year ago
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    @satellite73 , it cant be r=6 also? \[\frac{ x ^{2*(30-r)} }{ x ^{\frac{ r }{ 2 }} }=\frac{ x ^{2*(30-6)} }{ x ^{\frac{ 6 }{ 2 }} }=\frac{ x ^{48} }{ x ^{3} }\] \[x ^{48-3}=x ^{45}\] `?

  32. anonymous
    • one year ago
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    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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