1. ## Binomial Theorem problem

Hi, i have been trying to figure the below problem for ages, i can't figure it out, if someone could give me a lead that would very much appreciated

In the expansion (a+2bx)^12, a and b are constants. The coefficient of is 3168, while that of x^12 is . Find the values of a and b.

kleyzam

2. Originally Posted by kleyzam
Hi, i have been trying to figure the below problem for ages, i can't figure it out, if someone could give me a lead that would very much appreciated

In the expansion (a+2bx)^12, a and b are constants. The coefficient of is 3168, while that of x^12 is . Find the values of a and b.

kleyzam
The term with $x^{12}$ is $(2bx)^{12}$, so:

$(2b)^{12}=\frac{1}{4096}$

(and you should recognise $4096$ as $2^{12}$ )

The term with x^5 is:

$
{12 \choose 5}(2bx)^5a^7
$

so:

$
{12 \choose 5}(2b)^5a^7=3168
$

CB

3. I still can't figure out how to find the a and b

4. Originally Posted by CaptainBlack
$(2b)^{12}=\frac{1}{4096}$

(and you should recognise $4096$ as $2^{12}$ )
Originally Posted by kleyzam
I still can't figure out how to find the a and b
What have you done to try and solve the above equation for $b$?

CB