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! Thanks in advance! 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.
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
This has actually confused me even more :huh: The coefficient of x in my original question is 326592. But anyway, that is not what I'm asking about. 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.
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
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 I know they are both equivalent, but the fraction confuses me. So it would help me a lot if I understood your method. Thank's for your help!
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)
Hi haditya, 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!
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) Starting with your expression (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
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