Question

Trying to figure out the coefficient of x^5 in the expantion of (1+2x)^9. Can't seem to get the right answer...

Answer 1

You know something about Pascal's triangle? That would help you. :)

Answer 2

Yeah I tried (a+b)^n=nk=0n!k!(n−k)! akbn−k.and ended up with 126, when the answer sheet said 4032... I think I am misunderstanding the question

Answer 3

May I see your output?

Answer 4

I got \[9!\div4!.5!\] giving the answer 126

Answer 5

sorry didn't realise the time, i have to go to work. thanks for your help.

Answer 6

see https://en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem
for the details

the expansion of (2x+1)^9 contains the term \(a\ x^5\), defined to be
\[\left(\begin{matrix}9 \\ 4\end{matrix}\right)(2x)^{9-4} 1^4=126\cdot 2^5\cdot x^5=126\cdot 32\cdot x^5 = 4032 x^5\]

so you did 9 choose 4 correctly. But don't forget the coefficient on x (the 2) will also show up...