Question 48332
<pre><font size = 5><b>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/x<sup>2</sup>)<sup>12</sup>

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/x<sup>2</sup>)<sup>12</sup>

which contains x<sup>0</sup> which is 1 and that will make the term
be independent of x

We write 1/x<sup>2</sup> as x<sup>-2</sup>

The (r+1)st term of the expansion of (A + B)<sup>n</sup> is

        C(n,r)A<sup>r</sup>B<sup>n-r</sup>

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

        C(12,r)(2x)<sup>r</sup>(x<sup>-2</sup>)<sup>12-r</sup>

        C(12,r)2<sup>r</sup>x<sup>r</sup>x<sup>-2(12-r)</sup>

        C(12,r)2<sup>r</sup>x<sup>r</sup>x<sup>-24+2r</sup>

        C(12,r)2<sup>r</sup>x<sup>r-24+2r</sup>

        C(12,r)2<sup>r</sup>x<sup>3r-24</sup>

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)2<sup>r</sup>x<sup>3</sup>r<sup>-24</sup>

        C(12,8)2<sup>8</sup>x<sup>3(8)-24</sup>

        C(12,8)2<sup>8</sup>x<sup>0</sup>

        C(12,8)2<sup>8</sup>

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

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

And since 2<sup>8</sup> = 256, the answer is

        (495)(256) or 126720

Edwin</pre>