# binomial coefficient.

• May 12th 2010, 09:46 PM
brumby_3
binomial coefficient.
So I'm trying to find x^13 and x^14 in (4x^3-(1/x))^24

I've done:
(24)
(k) (4x^3)^K (-1/x)^(24-K)

(24)
(k) 4^Kx^3K (-1)^(24-k) x^(K-24)

(24)
(k) 4^K (-1)^(24-K) x^(3K+K-24)

(24)
(K) 4^K (-1)^(24-K) x^(4K-24)

and so for the 13th coefficient, 13 = 4K-24

PS at the start of each line the 24 and the K are together (ie in two brackets, I'm sure you'll get it)
• May 12th 2010, 10:16 PM
pickslides
Quote:

Originally Posted by brumby_3
which is 4K=37, but there's no integer answer for K? I'm sure this is wrong,

from your working this answer is fine, if $k \notin \mathbb{Z }$ it means there is no $x^{13}$ term
• May 12th 2010, 10:20 PM
brumby_3
Thanks!
For x^14:

14 = 4K-24
4K = 38
and that doesn't have an integer answer either. Is this right? So both x^13 and x^14 do not appear in the expansion, right? (From my lecture notes, the lecturer said there is no coefficient if it's not an integer answer.)
• May 12th 2010, 10:30 PM
pickslides
Looks like your lecturer and I agree. Best way to know for sure is to expand it out. If you have a good calculator or a computer package like Maple, it won't take long at all.