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 one year ago
In the expansion of (2a − 1)^n, the coefficient of the second term is −192. Find the value of n.
 one year 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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Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1I've gotten pretty far for this question but I just need a little help with the rest

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1\[\left(\begin{matrix}n \\ 1\end{matrix}\right)(2a)^{n1}(1)^1=192a^{n1}\]

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1\[(\frac{ n! }{ (n1)!1! })(2a)^{n1}=192a^{n1}\] \[\frac{ n! }{ (n1)! }=\frac{ 192a^{n1} }{ 2a^{n1} }\] \[\frac{ n! }{ (n1)! }=192\]

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1from here I do not know what to do

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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1also, \(n! = n (n1)!\)

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1WELCOME SOOO SOOO MUCH!!! ^_^

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1I am up to here:\[2^{n1}=\frac{ 192 }{ n }\]and I don't know what to do... Someone please help!

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

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1I know what the answer is, but I need to know how to get there

klimenkov
 one year 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.

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1now I am up to here: \[2^n=\frac{ 384 }{ n }\]

klimenkov
 one year 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.

sami21
 one year 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 ???

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1no, but I know that the answer is 6

sami21
 one year ago
Best ResponseYou've already chosen the best response.0ok newton method is a numerical technique used to solve equations . you familiar with derivatives ?

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1a little, I mean I am only an 11th grader

sami21
 one year 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 .

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1sure, thank you anyways for your help :)
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