Question

(x^2+1/x)的5次方展开式中x的系数是

Answer 1

第r+1项是T(r+1)=C5(r)*(x^2)^(5-r)*(1/x)^r=C5(r)*x^(10-2r-r)

令10-3r=1,则有r=3

即X的系数是C5(3)=10