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anonymous
 3 years ago
In the expansion of (2a − 1)^n, the coefficient of the second term is −192. Find the value of n.
anonymous
 3 years ago
In the expansion of (2a − 1)^n, the coefficient of the second term is −192. Find the value of n.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I've gotten pretty far for this question but I just need a little help with the rest

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}n \\ 1\end{matrix}\right)(2a)^{n1}(1)^1=192a^{n1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(\frac{ n! }{ (n1)!1! })(2a)^{n1}=192a^{n1}\] \[\frac{ n! }{ (n1)! }=\frac{ 192a^{n1} }{ 2a^{n1} }\] \[\frac{ n! }{ (n1)! }=192\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from here I do not know what to do

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\(\large (2a)^{n1}= 2^{n1}a^{n1}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, 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
 3 years ago
Best ResponseYou've already chosen the best response.1WELCOME SOOO SOOO MUCH!!! ^_^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am up to here:\[2^{n1}=\frac{ 192 }{ n }\]and I don't know what to do... Someone please help!

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0You can just put \(n=1,2,3,\ldots\) until you get the equality.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know what the answer is, but I need to know how to get there

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0It 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
 3 years ago
Best ResponseYou've already chosen the best response.0now I am up to here: \[2^n=\frac{ 384 }{ n }\]

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0This is a transcendential equation. There is no method to solve it using elementary operations and functions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well using newton method i am getting n=5.999 are you familiar with newton method ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, but I know that the answer is 6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok newton method is a numerical technique used to solve equations . you familiar with derivatives ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a little, I mean I am only an 11th grader

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm then i would say go for hit and trail .. start with n=0 ,1,2... untill you get the result .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure, thank you anyways for your help :)
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