binomial coefficient.

Jul 2008
212
2
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
which is 4K=37, but there's no integer answer for K? I'm sure this is wrong, please help!

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)
 

pickslides

MHF Helper
Sep 2008
5,237
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Melbourne
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 \(\displaystyle k \notin \mathbb{Z }\) it means there is no \(\displaystyle x^{13}\) term
 
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Jul 2008
212
2
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.)
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
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.
 
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