Question

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

Answer 1

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

Answer 2

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

Answer 3

\[-(\frac{ n! }{ (n-1)!1! })(2a)^{n-1}=-192a^{n-1}\]
\[-\frac{ n! }{ (n-1)! }=-\frac{ 192a^{n-1} }{ 2a^{n-1} }\]
\[\frac{ n! }{ (n-1)! }=192\]

Answer 4

from here I do not know what to do

Answer 5

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

Answer 6

oh, oops!

Answer 7

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

Answer 8

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!!!

Answer 9

WELCOME SOOO SOOO MUCH!!! ^_^

Answer 10

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

Someone please help!

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

This is a transcendential equation. There is no method to solve it using elementary operations and functions.

Answer 16

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

Answer 17

no, but I know that the answer is 6

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

sure, thank you anyways for your help :)