# Thread: Finding a Binomial Coefficient

1. ## Finding a Binomial Coefficient

What is the coefficient of x^5 in the binomial expression (x - 2x^-2)^20 when multiplied by (1+3x^3)

So far I have:
(20 over k)(-1)^k (2)^k x^(20-3k)

so then 20-3k = 5 such that k = 5

So then (20 over 5 )(-1)^5 (2)^5

But then I don't know how to handle multiplying it by ( 1 + 3x^3)

2. ## Re: Finding a Binomial Coefficient

$(x - 2x^{-2})^{20}=a_{20}x^{20}+a_{17}x^{17}+\dots+a_5x^5+a_ 2x^2+\dots+a_{-37}x^{37}+a_{-40}x^{-40}$ for some coefficients $a_k$. Also, $1+3x^3=b_0x^0+b_3x^3$ for $b_0=1$, $b_3=3$. The coefficient of $x^5$ in $(x - 2x^{-2})^{20}(1+3x^3)$ is $\sum_{i+k=5}a_ib_k$, i.e., $a_2b_3+a_5b_0$.

3. ## Re: Finding a Binomial Coefficient

Originally Posted by ThatPinkSock
What is the coefficient of x^5 in the binomial expression (x - 2x^-2)^20 when multiplied by (1+3x^3)
So far I have: (20 over k)(-1)^k (2)^k x^(20-3k)
so then 20-3k = 5 such that k = 5
So then (20 over 5 )(-1)^5 (2)^5
but then I don't know how to handle multiplying it by ( 1 + 3x^3)
I have a different take on this question from you and reply #2.
Each term in $\left(x-2x^{-2}\right)^{20}$looks like $\binom{20}{k}(-2)^{20-k}x^{3k-40}$
Now you want $x^5$ from $[x^{3k-40}]~\&~3x^3{x^{3k-40}}$
So $k=14~\&~k=15$ WHY?