anonymous
  • anonymous
In the expansion of (2a − 1)^n, the coefficient of the second term is −192. Find the value of n.
Mathematics
schrodinger
  • schrodinger
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schrodinger
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anonymous
  • anonymous
I've gotten pretty far for this question but I just need a little help with the rest
anonymous
  • anonymous
\[\left(\begin{matrix}n \\ 1\end{matrix}\right)(2a)^{n-1}(-1)^1=-192a^{n-1}\]
anonymous
  • anonymous
\[-(\frac{ n! }{ (n-1)!1! })(2a)^{n-1}=-192a^{n-1}\] \[-\frac{ n! }{ (n-1)! }=-\frac{ 192a^{n-1} }{ 2a^{n-1} }\] \[\frac{ n! }{ (n-1)! }=192\]

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anonymous
  • anonymous
from here I do not know what to do
hartnn
  • hartnn
\(\large (2a)^{n-1}= 2^{n-1}a^{n-1}\)
anonymous
  • anonymous
oh, oops!
hartnn
  • hartnn
also, \(n! = n (n-1)!\)
anonymous
  • anonymous
oh, right I think I can solve the rest by myself. I'll ask you if I need any more help... THANK YOU SOOO SOOO MUCH!!!
hartnn
  • hartnn
WELCOME SOOO SOOO MUCH!!! ^_^
anonymous
  • anonymous
I am up to here:\[2^{n-1}=\frac{ 192 }{ n }\]and I don't know what to do... Someone please help!
klimenkov
  • klimenkov
You can just put \(n=1,2,3,\ldots\) until you get the equality.
anonymous
  • anonymous
I know what the answer is, but I need to know how to get there
klimenkov
  • klimenkov
It cant be solved so easy. The way to solve is just to put the values of n into the equation and watch what you get.
anonymous
  • anonymous
now I am up to here: \[2^n=\frac{ 384 }{ n }\]
klimenkov
  • klimenkov
This is a transcendential equation. There is no method to solve it using elementary operations and functions.
anonymous
  • anonymous
well using newton method i am getting n=5.999 are you familiar with newton method ???
anonymous
  • anonymous
no, but I know that the answer is 6
anonymous
  • anonymous
ok newton method is a numerical technique used to solve equations . you familiar with derivatives ?
anonymous
  • anonymous
a little, I mean I am only an 11th grader
anonymous
  • anonymous
hmm then i would say go for hit and trail .. start with n=0 ,1,2... untill you get the result .
anonymous
  • anonymous
sure, thank you anyways for your help :)

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