Posted by anoynomous on Friday, March 22, 2013 at 1:17pm.
coefficient of x^39 in the expansion of [(1/x^2) + x^4)]^18
maths-binomial theorem - Reiny, Friday, March 22, 2013 at 2:15pm
Use the general term
term(n+1) = C(18,n) (1/x^2)^(18-n) (x^4)^n
= C(18,n) (x^(-36 +2n) ) (x^4n)
= c(18,n) x^(2n - 36)
ahhh, trick question,
to have a term with x^39 , 2n-36 = 39
there is no whole number solution,
so there is no term with x^39
e.g. --- take the first few terms
[(1/x^2) + x^4)]^18
= (1/x^2)^18 + 18(1/x^2)^17 (x^4) + 153 (1/x^2)^16 (x^4)^2 + ..
= x^-36 + 18x^-30 + 153x^-24 + ..
the exponent go
-36 -30 -24 .. missing the 39
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