已知对于数列{an}中,有fn(x)=a1x+a2x^2+...+anx^n,且a1=3,fn(1)=p*(2^n-1/

fn(1)=p*(2^n-1/2^n)=Sn an=Sn-S(n-1)=p*(2^n-1/2^n)-p*[2^(n-1)-1/2^(n-1)]
=p{[2^n-2^(n-1)]-[1/2^n-1/2^(n-1)]}=p{2^(n-1)+1/2^n}
an=p*2^(n-1)+p/2^n
以上回答你满意么?