mathsbinomial theorem
posted by anoynomous .
coefficient of x^39 in the expansion of [(1/x^2) + x^4)]^18

Use the general term
term(n+1) = C(18,n) (1/x^2)^(18n) (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 , 2n36 = 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