# Binomial Expansion Question

• Apr 24th 2008, 06:30 PM
K.Chopin
Binomial Expansion Question
Hi everyone! I am new to the forum, and happy(Rofl) I found it. So, I'm almost finished with college algebra (just one more test before my final) and wouldn't you know it, this section on binomial expansion is not going very well!
Anyway, this is the question on my review sheet, "What is the coefficient on the X^6Y^5 term in the expansion of (X^2+2Y)^8?"
I used "n above k" and thought the answer was 28; however, I was way off. What did I forget to do? I think it has something to do with the 2Y part, but I'm stuck.
If you can offer any help, I would greatly appreciate it!

Thanks,
Kay
• Apr 24th 2008, 06:44 PM
o_O
$\displaystyle \left(x^{2} + 2y\right)^{8} = {8 \choose 0}\left(x^{2}\right)^{8} + {8 \choose 1}\left(x^{2}\right)^{7}(2y) + ... + {8 \choose 7}\left(x^{2}\right)^{1}(2y)^{7} + {8 \choose 8}(2y)^{8}$

What you're looking for is:
$\displaystyle {8 \choose 5}\left(x^{2}\right)^{3}(2y)^{5} = {8 \choose 5}x^{6}(2y)^{5}$ (here's your x^6y^5)

So you can see that the coefficient is going to be: $\displaystyle {8 \choose 5} \cdot 2^{5}$