In the binomial expansion of (a+b)^12 determine

A) the value of k in the term containing a^2k b^k
B) the coefficient of that term, first in c(n,r) form, and then evaluate
C) the number of terms in the expansion
D) the middle term

I don't understand
This is from my statistics/ data management class thanks

This looks to be a pretty complete explanation of what's involved. Look it up and if you still have problems, post back.

Binomial theorem - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Binomial_theorem)

I understand the expansion I just don't get how to solve this problem because it looks to be different

It's not different.

You can do it the brute-force way using Pascal's triangle. Keep going until the second number = 12.
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

But it gets a bit long at 12.

You can use equation 1 on the Wikipedia page:

Binomial theorem - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Binomial_theorem)

to derive any term including the coefficient. That's all you're asked to do.

For (C); Or you could try seeing the pattern, quadratics have 3 terms

(a+b)^2 = a^2 + 2ab + b^2

Cubics four terms,

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

Do you have a nice calculator? Just use that for an easy way out.

Expand (a+b)^{12}

There is a distinct pattern to these. Let me show you a trick to finding the coefficients without all that combinatorics stuff. OK?

The first term is a^{12}b^{0}. Since b^{0}=1, we do not bother writing it.

To find the next coefficient, it is always the power term.

So, subtract 1 from the 'a' power and add 1 to the 'b' power.

The next term is then 12a^{11}b

Now, to find the next coefficient, multiply 12 by the 'a' power and then divide by the 'b' power plus 1.

We get (12*11)/2=66

Therefore, the next term is 66a^{10}b^{2}

The next term would be (66*10)/3=220, 220a^{9}b^{3}

The next coefficient would be (220*9)/4=496

Continue in that manner until you get to a coefficient of 924, then they repeat back down to the end at b^12.

See?

How many terms are there? Well, one more than the power of given binomial. Wouldn't that be 13?