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I've gotten pretty far for this question but I just need a little help with the rest

\[\left(\begin{matrix}n \\ 1\end{matrix}\right)(2a)^{n-1}(-1)^1=-192a^{n-1}\]

from here I do not know what to do

\(\large (2a)^{n-1}= 2^{n-1}a^{n-1}\)

oh, oops!

also, \(n! = n (n-1)!\)

WELCOME SOOO SOOO MUCH!!! ^_^

I am up to here:\[2^{n-1}=\frac{ 192 }{ n }\]and I don't know what to do...
Someone please help!

You can just put \(n=1,2,3,\ldots\) until you get the equality.

I know what the answer is, but I need to know how to get there

now I am up to here:
\[2^n=\frac{ 384 }{ n }\]

well using newton method i am getting n=5.999
are you familiar with newton method ???

no, but I know that the answer is 6

ok newton method is a numerical technique used to solve equations .
you familiar with derivatives ?

a little, I mean I am only an 11th grader

hmm then i would say go for hit and trail ..
start with n=0 ,1,2... untill you get the result .

sure, thank you anyways for your help :)