# Simple Fractions Question

#### TheTechGod

Joined Jun 15, 2005
8
I'm studying for a test which will contain questions of how to obtain binomial coefficients, I understand the idea and can do it when the equation does not contain fractions but have trouble when it does contain fractions.

How does one go from:

(2x)^(9-k) (3/x)^k

to...

2^(9-k) 3^k x^(9-2k)

???

I'm sure it's an easy year 8 level problem, But I have not worked with fractions for a long time and seem to have forgotten it! PS:
For anyone actually interested in the binomial question, it is as follows:

Find the coefficient of x in the expansion of (2x + 3/x)^9

I have the answer if anyone wants it, I just don't understand the fractions part which I asked about above.

Joined Jan 19, 2004
220
it is equivalent to finding the coeff of x^10 in the expansion of (2*x*x+3)^9

this involves getting the 1/x term out of the expansion to the foll:
x^-9(2*x*x+3)^9

put x*x=y
exp=(2*y+3)^9

req answer= 9C5*(2^5)*(3^4)...sorry i goofed up here

#### TheTechGod

Joined Jun 15, 2005
8
Originally posted by haditya@Sep 11 2005, 02:03 AM
it is equivalent to finding the coeff of x^10 in the expansion of (2*x*x+3)^9

this involves getting the 1/x term out of the expansion to the foll:
x^-9(2*x*x+3)^9

put x*x=y
exp=(2*y+3)^9

[post=10235]Quoted post[/post]​
This has actually confused me even more :huh:

The coefficient of x in my original question is 326592.

The full question with the worked out answer is:

Find the coefficient of x in the expansion of (2x + 3/x)^9

For this expression, a general term is:

(9 C K) (2x)^(9 - k) (3/x)^k

This can be rewritten as

(9 C K) 2^(9 - k) 3^k x^(9 - 2k)

To obtain a exponent of 1, we require 9 - 2k = 1, or k = 4. Then we have:

(9 C 4) (2^5) (3^4) = 126(32)(81) = 326592.

Now, I understand all that... except the part where (2x)^(9 - k) (3/x)^k changed to 2^(9 - k) 3^k x^(9 - 2k).

I understand why it was changed, but not how it was done.

I also realise that this is a simple year 8 fractions problem, and that I should know it!

Can anyone help?

Thanks.

#### hgmjr

Joined Jan 28, 2005
9,029
Originally posted by haditya@Sep 10 2005, 11:03 AM
it is equivalent to finding the coeff of x^10 in the expansion of (2*x*x+3)^9

this involves getting the 1/x term out of the expansion to the foll:
x^-9(2*x*x+3)^9

put x*x=y
exp=(2*y+3)^9

[post=10235]Quoted post[/post]​
I'm guessing that haditya meant to write
As it turns out C(9,5) = C(9,4) so you both arrived at the same answer.

Using the Binomial Series method I came up with the general espression for the set of coefficients associated with the given equation.

C(9,K)*(2)^K*(x)^K*(3)^(9-K)*(x)^-(9-K)

Taking the exponents of x in the above expression and summing them together and then setting them equal to 1 we get:

K+ -(9-K) = 1
2(K)-9 = 1
2(K) = 10
K = 5

Plug in K=5 in the the general expression for the set of coefficents and I get

C(9,5)(2^5)(x^5)(3^4)(x^-4)

126*32*81*x = 326592x

hgmjr

#### TheTechGod

Joined Jun 15, 2005
8
Hi hgmjr,

Using the Binomial Series method I came up with the general espression for the set of coefficients associated with the given equation.

C(9,K)*(2)^K*(x)^K*(3)^(9-K)*(x)^-(9-K)
Can you explain how you got this? The way we have been taught, gives us the following:

(9 C K) (2x)^(9 - k) (3/x)^k

I know they are both equivalent, but the fraction confuses me. So it would help me a lot if I understood your method.

Joined Jan 19, 2004
220
oh ok tha law of incides u mean
1. a^m*a^n=a^(m+n)
2. a^m/a^n=a^(m-n)
3. (a*B)^m= a^m*b^m
4. (a^m)^n=a^(mn)

#### TheTechGod

Joined Jun 15, 2005
8

I think I actually understand it now. I was digging up stuff on fractions when I should have been looking for indices.

So...

(2x)^(9 - k) (3/x)^k

becomes

2^(9 - k) 3^k

Since the x's cancel...

I'm still not sure why the teachers notes show it as:

(2x)^(9 - k) (3/x)^k

becomes

2^(9 - k) 3^k x^(9 - 2k)

But at least I can get the right answer now...

Thank's all!

#### hgmjr

Joined Jan 28, 2005
9,029
Originally posted by TheTechGod@Sep 11 2005, 01:23 AM
Hi hgmjr,
Can you explain how you got this? The way we have been taught, gives us the following:

(9 C K) (2x)^(9 - k) (3/x)^k

[post=10249]Quoted post[/post]​
Hi TheTechGod,

The set of identities for manipulating exponents that haditya has posted contains the ones I used in arriving at my expression for the general case for the set of coefficients for your specific problem statement.

1. a^m*a^n=a^(m+n)
2. a^m/a^n=a^(m-n)
3. (a*b )^m= a^m*b^m
4. (a^m)^n=a^(mn)

(9 C K) (2x)^(9 - k) (3/x)^k

It can be expanded using identity 3 above to get

(9CK)*(2)^(9-K)*(x)^(9-K)*(3/x)^(K)

It can be further expanded using identity 2 above to get

(9CK)*(2)^(9-K)*(x)^(9-K)*(3)^(K)*(x)^(-K)

Taking the exponents of the two x terms above.

(9 - K)+(-K) = 1

9-2K = 1
-2K = -8
K=4

Then plugging K=4 the following equation is obtained

(9C4)*(2)^(5)*(x)^(5)*(3)^(4)*(x)^(-4)

Collecting the two x terms yields

(9C4)*(2^5)*(3^4)*(x^5)*(x^-4)

Rearranging and then combining the two x terms using identity 1 yields

(9C4)*(2^5)*(3^4)*(x)

As I mention in my earlier reply (9C4) evaluates to the same value as (9C5) so that choice of which convention to use between (9-K),(K) or (K),(9-K) yields the same result.

Hope this helps.

hgmjr

Joined Jan 19, 2004
220
the x's dont cancel but add up.
coz its (1/x)^k * x^(9-k)
1/x=x^-1
therefore the result becomes (x^-1)^k * x^(9-k)
(x^-k) * x^(9-k)

hence the result

Thanks for everything, I passed the test 