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

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Chelsea04Best ResponseYou've already chosen the best response.1
I've gotten pretty far for this question but I just need a little help with the rest
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
\[\left(\begin{matrix}n \\ 1\end{matrix}\right)(2a)^{n1}(1)^1=192a^{n1}\]
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
\[(\frac{ n! }{ (n1)!1! })(2a)^{n1}=192a^{n1}\] \[\frac{ n! }{ (n1)! }=\frac{ 192a^{n1} }{ 2a^{n1} }\] \[\frac{ n! }{ (n1)! }=192\]
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
from here I do not know what to do
 one year ago

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

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

Chelsea04Best ResponseYou've already chosen the best response.1
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!!!
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
WELCOME SOOO SOOO MUCH!!! ^_^
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
I am up to here:\[2^{n1}=\frac{ 192 }{ n }\]and I don't know what to do... Someone please help!
 one year ago

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

Chelsea04Best ResponseYou've already chosen the best response.1
I know what the answer is, but I need to know how to get there
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
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.
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
now I am up to here: \[2^n=\frac{ 384 }{ n }\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
This is a transcendential equation. There is no method to solve it using elementary operations and functions.
 one year ago

sami21Best ResponseYou've already chosen the best response.0
well using newton method i am getting n=5.999 are you familiar with newton method ???
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
no, but I know that the answer is 6
 one year ago

sami21Best ResponseYou've already chosen the best response.0
ok newton method is a numerical technique used to solve equations . you familiar with derivatives ?
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
a little, I mean I am only an 11th grader
 one year ago

sami21Best ResponseYou've already chosen the best response.0
hmm then i would say go for hit and trail .. start with n=0 ,1,2... untill you get the result .
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
sure, thank you anyways for your help :)
 one year ago
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