3.先对Pn(x)/(x^k-1)^(n+1)求导
得到Pn'(x)/(x^k-1)^(n+1)-(n+1)*k*x^(k-1)*Pn(x)/(x^k-1)^(n+2)
可知Pn+1(x)=Pn'(x)*(x^k-1)-(n+1)*k*x^(k-1)*Pn(x)
所以Pn+1(1)=-(n+1)*k*Pn(1)
根据递推数列特征:Pn(1)=(-k)^(n-1)*n!*P1(1)
然而对1/(x^k-1)求导
得到-k*x^(k-1)/(x^k-1)^2
所以P1(1)=-k
于是,Pn(1)=(-k)^(n-1)*n!*P1(1)=(-k)^n*n!*
第一题还没想出来.