Question

How to approach this?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

you have ignored the second term in your expansion

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yeah exactly

Answer 9

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

Answer 10

24?

Answer 11

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

Answer 12

no im not quite following

Answer 13

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

Answer 14

i am off by a minus sign maybe

Answer 15

oh no really

Answer 16

now it is clear that it is a constant?

Answer 17

yeah cause the x:es = 1?

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

right

Answer 22

now there may be another one as well

Answer 23

oh no maybe not

Answer 24

oh yes i changed my mind i think there is
try \(r=6\)

Answer 25

you still get an exponent of 12 top and bottom

Answer 26

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

Answer 27

yeah exactly

Answer 28

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

Answer 29

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

Answer 30

2 * 593775

Answer 31

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

`?

Answer 32

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