正项数列{an}的前n项和Sn=1/4an^2+1/2an-3/4,且a1b1+a2b2+a3b3+…+anbna1b1+a2b2+a3b3+…+anbn=2的(n+1)次方+2,求数列{bn}的通项公式
问题描述:
正项数列{an}的前n项和Sn=1/4an^2+1/2an-3/4,且a1b1+a2b2+a3b3+…+anbn
a1b1+a2b2+a3b3+…+anbn=2的(n+1)次方+2,求数列{bn}的通项公式
答
4Sn=(An)^2+2An-34A1=4S1=(A1)^2+2A1-3(A1)^2-2A1-3=0(A1-3)(A1+1)=0A1=3 A1=-1(舍去)4An=4Sn-4S(n-1)=((An)^2+2An-3)-((A(n-1))^2+2A(n-1)-3)(An)^2-(A(n-1))^2=2An+2A(n-1)(An+A(n-1))(An-A(n-1))=2(An+A(n-1))An>...