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