## Chelsea04 2 years ago In the expansion of (2a − 1)^n, the coefficient of the second term is −192. Find the value of n.

1. Chelsea04

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

2. Chelsea04

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

3. Chelsea04

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

4. Chelsea04

from here I do not know what to do

5. hartnn

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

6. Chelsea04

oh, oops!

7. hartnn

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

8. Chelsea04

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!!!

9. hartnn

WELCOME SOOO SOOO MUCH!!! ^_^

10. Chelsea04

I am up to here:$2^{n-1}=\frac{ 192 }{ n }$and I don't know what to do... Someone please help!

11. klimenkov

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

12. Chelsea04

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

13. 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.

14. Chelsea04

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

15. klimenkov

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

16. sami-21

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

17. Chelsea04

no, but I know that the answer is 6

18. sami-21

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

19. Chelsea04

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

20. sami-21

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

21. Chelsea04

sure, thank you anyways for your help :)

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