题目

数列{an}满足a(n+1)+an=4n-3,若{an}是等差数列,(1)求{an}的通项公式(2)设Sn是{an}的前n项和,
数列{an}满足a(n+1)+an=4n-3,若{an}是等差数列,(1)求{an}的通项公式
(2)设Sn是{an}的前n项和,且a1=1,求S（2n+1）

提问时间：2021-02-26

答案

（1）a(n+1)+an=4n-3
{an}是等差数列→a(n+1)-an=d
两式相加,a(n+1)=(4n+d-3)/2=2n+(d-3)/2
∴an=2(n-1)+(d-3)/2=2n+(d-3)/2-2=2n+(d-7)/2
两式相减,an=(4n-d-3)/2=2n-(d+3)/2
∴(d-7)/2=-(d+3)/2,d=2
∴an=2n-(d+3)/2=2n-5/2
（2）（a1明明等于-1/2,所以应去掉“{an}是等差数列”的条件）
S(2n+1)
=a1+(a2+a3)+(a4+a5)+...+a(2n)+a(2n+1)
=1+(4*2-3)+(4*4-3)+...+(4*2n-3)
=(8*1+8*2+...+8*n)+1-3(n-1)
=8(1+2+...+n)+1-3n-3
=8*n(n+1)/2-3n-2
=4n(n+1)-3n-2
=4n^2+4n-3n-3
=4n^2+n-3