Thread: NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

1. NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

Going through past paper for the above question to find the coefficient of the above expansion & have seen a guide to using my Casio here to work out the value of coefficient at x^14 but it errors for this calculation.
n should be bigger than r, shouldn't it?
i.e. n=14 & r=16
nCr = n!/(n-r)!r! which is 14! / -2! 16! so no suprise there, perhaps the indeces are wrong?

So, next example seems more likely; for the coefficient of x^18 for expansion of (1/14 x^2 - 7)^16.
so n=18 & r=16

nCr = 18! / -2! 16! = 153

I have a feeling there is something else, but not sure what it is.
Could someone point in the right direction

2. Re: NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

Originally Posted by froodles01
Going the value of coefficient at x^14 but it errors for this calculation. n should be bigger than r, shouldn't it?
i.e. n=14 & r=16
nCr = n!/(n-r)!r! which is 14! / -2! 16! so no suprise there, perhaps the indeces are wrong?
You want $\displaystyle x^{14}$. So $\displaystyle \binom{16}{2}(3x)^{14}\left(\frac{-1}{3}\right)^2$

3. Re: NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

Thank you, the 3x term is raised to 14 for the coefficient of that term, but I don't understand why the 1/3 term is squared not raised to 14 too.

Can you apply this to the second example, too.
Thank you.

4. Re: NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

Originally Posted by froodles01
Thank you, the 3x term is raised to 14 for the coefficient of that term, but I don't understand why the 1/3 term is squared not raised to 14 too.
The terms of the expansion $\displaystyle (x+y)^N$ are $\displaystyle \binom{N}{k}x^ky^{N-k}$.
In your question $\displaystyle x=3x,~y=\tfrac{-3}{x}~\&~N=16$ .

5. Re: NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

Thank you. I'm getting there, but in the second example where N=16, the k value is -2

Should this look like: (16 -2) (1/4 x^2)^18 (-7)^-2
= 4.322exp-24

6. Re: NCr (nCk) find binomial coefficient of x^14 for expansion of (3x - 1/3)^16?

Originally Posted by froodles01
So, next example seems more likely; for the coefficient of x^18 for expansion of (1/14 x^2 - 7)^16.
In this one, $\displaystyle N=16,~k=9~\&~(N-k)=7$.