Find the numerical value of the coefficient x^11 in the expansion of (x^2+1/x)^10 in the powers of x.

Answer 1

general term = term(r+1)
=C(10,r) (x^2)^(10-r) (1/x)^r
= C(10,r) x^(20 - 2r) x^(-r)
= C(10,r) x^(20 -3r)

so 20 -3r = 11
-3r = -9
r = 3

C(10,3) = 120
the term containing x^11 is 120x^11 and it is the 4th term

confirmation by Wolfram:
http://www.wolframalpha.com/input/?i=expand+%28x%5E2%2B1%2Fx%29%5E10