来自程葳的问题
正项数列{an}满足:an2-(2n-1)an-2n=0.(1)求数列{an}的通项公式an;(2)令bn=1(n+1)an,求数列{bn}的前n项和Tn.
正项数列{an}满足:an2-(2n-1)an-2n=0.
(1)求数列{an}的通项公式an;
(2)令bn=1(n+1)an,求数列{bn}的前n项和Tn.
1回答
2020-01-16 19:03
正项数列{an}满足:an2-(2n-1)an-2n=0.(1)求数列{an}的通项公式an;(2)令bn=1(n+1)an,求数列{bn}的前n项和Tn.
正项数列{an}满足:an2-(2n-1)an-2n=0.
(1)求数列{an}的通项公式an;
(2)令bn=1(n+1)an,求数列{bn}的前n项和Tn.
(1)由正项数列{an}满足:a2n-(2n-1)an-2n=0,可得(an-2n)(an+1)=0所以an=2n.(2)因为an=2n,bn=1(n+1)an,所以bn=1(n+1)an=12n(n+1)=12(1n−1n+1),Tn=12(1−12+12−13+…+1n−1n+1)=12(1−1n+1)=n2n+2....