Question

is there a faster way of calculating the coefficients for x^2, x^3, x^4 than fully expanding ((2/x)+3x)^5

Answer 1

Binomial expansion!

Answer 2

no, I know that technique, I was just wondering if there was a faster way...or is that the fastest?

Answer 3

\[\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)x^{k}y^{n-k}\]

Answer 4

You dont' need to fully expand it.

Or, just the pascals triangle; but its stupiiidddd!

Answer 5

oh, ok
good to know
Thank-you for your help :)