- #1

mr_coffee

- 1,629

- 1

Hello everyone. I was wondering if there was a fast way to figure out the following:

THe question is:

What is the cofficient of u^16v^2 in the binomial expansion of (u^2 + v)^10?

Well the answer is 45.

I know what the binomail theorem is:

and I can put (u^2 + v)^10 in that form

a = u^2

b = v

n = 10

(10 choose 0)*(u^2)^10 + (10 choose 1)*(u^2)^(9)*v + (10 choose 2)*(u^2)^(8)*v^2...

Okay so i found (10 choose 2) is where the coefficent of the binomial expansion of (u^2 + v)^10. But is this how your supppose to do it? Write it out like that? Or is there a faster way?

I also got 45 by just taking (10 choose 2) = 45.

So my question is, is there another way to compute this? It looks like the professor did somthing like:

(u^16)*v^2 = (u^2)^8*v^2 I see they equal each other but I'm not sure how this is connecting things together and how its figuring out the co-efficent.

ANy help would be great!

THe question is:

What is the cofficient of u^16v^2 in the binomial expansion of (u^2 + v)^10?

Well the answer is 45.

I know what the binomail theorem is:

and I can put (u^2 + v)^10 in that form

a = u^2

b = v

n = 10

(10 choose 0)*(u^2)^10 + (10 choose 1)*(u^2)^(9)*v + (10 choose 2)*(u^2)^(8)*v^2...

Okay so i found (10 choose 2) is where the coefficent of the binomial expansion of (u^2 + v)^10. But is this how your supppose to do it? Write it out like that? Or is there a faster way?

I also got 45 by just taking (10 choose 2) = 45.

So my question is, is there another way to compute this? It looks like the professor did somthing like:

(u^16)*v^2 = (u^2)^8*v^2 I see they equal each other but I'm not sure how this is connecting things together and how its figuring out the co-efficent.

ANy help would be great!