Today my teacher informed us we would have a EPW test
coming up in a few days. We were given little information 
but told it would include questions such as this:

Q: Find the term independent of x:
(2x + 1/x2)12

It's probably linked to counting techniques as that's the 
topic we're currently studying. As far as i know it's not
from a txtbook.

Thanx for any help
AC, Aus.

We have to find the term in the binomial expansion of

(2x + 1/x2)12

which contains x0 which is 1 and that will make the term
be independent of x

We write 1/x2 as x-2

The (r+1)st term of the expansion of (A + B)n is

        C(n,r)ArBn-r

so the (r+1)st term of the expansion of (2x + x-2)12 is

        C(12,r)(2x)r(x-2)12-r

        C(12,r)2rxrx-2(12-r)

        C(12,r)2rxrx-24+2r

        C(12,r)2rxr-24+2r

        C(12,r)2rx3r-24

Now we require the power of x to be 0, so we set

             3r-24 = 0
                3r = 24
                 r = 8

So the (8+1)st or 9th term is the one free of x
Substituting 8 for r in
 
        C(12,r)2rx3r-24

        C(12,8)28x3(8)-24

        C(12,8)28x0

        C(12,8)28

The formula for C(n,r) is n!/[r!(n-r)]

So C(12,8) = 12!/[8!(8-2)!] = 495

And since 28 = 256, the answer is

        (495)(256) or 126720

Edwin