anonymous one year ago How to approach this?

1. anonymous

Examine if $(x ^{2}-\frac{ 1 }{ \sqrt{x} })^{30}$ contain a constant term and if so, decide it.

2. anonymous

it will be a number when you have the same exponent in the numerator and denominator

3. anonymous

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

4. anonymous

you have ignored the second term in your expansion

5. anonymous

like i said, should be a constant when the power in the numerator is equal to the power in the denominator

6. anonymous

power in the numerator will be $$2(30-k)$$ and in the denominator will be $$\frac{1}{2}k$$ or vice versa

7. anonymous

$=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}$

8. anonymous

yeah exactly

9. anonymous

so for what value of $$k$$ is $$2(30-k)=\frac{1}{2}k$$ or ]$$r$$ or whatever your variable is?

10. anonymous

24?

11. anonymous

yes there is another one as well, but is it clear why that will give a constant?

12. anonymous

no im not quite following

13. anonymous

$\binom{30}{24}\frac{x^{2\times 6}}{x^{\frac{24}{2}}}$

14. anonymous

i am off by a minus sign maybe

15. anonymous

oh no really

16. anonymous

now it is clear that it is a constant?

17. anonymous

yeah cause the x:es = 1?

18. anonymous

no because the exponent in the numerator and denominator are both 12

19. anonymous

ooh, yeah but that will be $x ^{12-12}=x^{0}=1$ ?

20. anonymous

$=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}(\frac{ 1 }{ \sqrt{x}^{?}})^{r}$ $=\sum_{r=0}^{30}\left(\begin{matrix}30 \\ r\end{matrix}\right)(x^2)^{30-r}\left(x^{-\frac{1}{2}}\right)^r$

21. anonymous

right

22. anonymous

now there may be another one as well

23. anonymous

oh no maybe not

24. anonymous

oh yes i changed my mind i think there is try $$r=6$$

25. anonymous

you still get an exponent of 12 top and bottom

26. anonymous

$\left(\begin{matrix}30 \\ 6\end{matrix}\right)$

27. anonymous

yeah exactly

28. anonymous

so, unless i am way off base, your answer is $2\times \binom{30}{6}$

29. anonymous

It is 2 x 30 choose 6 because 30 choose 6 and 30 choose 24 is the same?

30. anonymous

2 * 593775

31. anonymous

@satellite73 , it cant be r=6 also? $\frac{ x ^{2*(30-r)} }{ x ^{\frac{ r }{ 2 }} }=\frac{ x ^{2*(30-6)} }{ x ^{\frac{ 6 }{ 2 }} }=\frac{ x ^{48} }{ x ^{3} }$ $x ^{48-3}=x ^{45}$ `?

32. anonymous

but still $\left(\begin{matrix}30 \\ 24\end{matrix}\right) \leftarrow \rightarrow \left(\begin{matrix}30 \\ 6\end{matrix}\right)$ right?