作业帮已知数列{an}的前n项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n>=2),计算S1,S2,S3,S4,并猜想Sn表达式.
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/01 06:31:06
已知数列{an}的前n项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n>=2),计算S1,S2,S3,S4,并猜想Sn表达式.
已知数列{an}的前n项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n>=2),计算S1,S2,S3,S4,并猜想Sn表达式.
已知数列{an}的前n项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n>=2),计算S1,S2,S3,S4,并猜想Sn表达式.
S(1) = a(1) = -2/3
S(2) = S(1) + a(2) = a(2) - 2/3,a(2) = S(2) + 2/3
S(2) + 1/S(2) + 2 = a(2) = S(2) + 2/3,
1/S(2) = -4/3,
S(2) = -3/4
S(3) = S(2) + a(3) = a(3) - 3/4,a(3) = S(3) + 3/4
S(3) + 1/S(3) + 2 = a(3) = S(3) + 3/4,
1/S(3) = -5/4,
S(3) = -4/5.
S(4) = S(3) + a(4) = a(4) - 4/5,a(4) = S(4) + 4/5.
S(4) + 1/S(4) + 2 = a(4) = S(4) + 4/5,
1/S(4) = -6/5,
S(4) = -5/6.
猜想
S(n) = -(n+1)/(n+2),n = 1,2,...
---------
证明
n = 1时,S(1) = -2/3,结论成立.
设 n = k时,有S(k) = -(k+1)/(k+2)
当n = k+1时,
S(k+1) = S(k) + a(k+1) = a(k+1) - (k+1)/(k+2),
a(k+1) = S(k+1) + (k+1)/(k+2),
S(k+1) + 1/S(k+1) + 2 = a(k+1) = S(k+1) + (k+1)/(k+2),
1/S(k+1) = -(k+3)/(k+2),
S(k+1) = -(k+2)/(k+3),结论成立.
所以,
由归纳法,猜想正确.
楼上的方法很好嘛！归纳法虽说是大学的课程，但是判高考卷的都是大学老师，所以嘛。。。什么方法都行！