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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