数列 (2 14:21:46) 已知a1=2,an+1=(√2-1)(an +2)(n=1,2,3,……),则{an}的通项公式为an=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 04:16:05
![数列 (2 14:21:46) 已知a1=2,an+1=(√2-1)(an +2)(n=1,2,3,……),则{an}的通项公式为an=](/uploads/image/z/12054391-7-1.jpg?t=%E6%95%B0%E5%88%97+%282+14%3A21%3A46%29%26%23160%3B%E5%B7%B2%E7%9F%A5a1%3D2%2Can%2B1%3D%28%E2%88%9A2-1%29%EF%BC%88an+%2B2%EF%BC%89%28n%3D1%2C2%2C3%2C%E2%80%A6%E2%80%A6%29%2C%E5%88%99%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E4%B8%BAan%3D%26%23160%3B%26%23160%3B%26%23160%3B%26%23160%3B%26%23160%3B%26%23160%3B%26%23160%3B%26%23160%3B+%26%23160%3B)
xՒN@_eSҒ:#a| ށd| Ҙ`"ZL1)xA](Yȥw!x3rѸԶ挕I@C &bo^A6]g
t4
@ƶ1'b0zHwO&9wrc'nPTu&e~ +vL'ʝ6΄seV@)Q`YnٰR ;Rjx?T ,ScPV-rT; PI
.:f$DgthS12&?dEdJ"++_@`6QT¯Aw/OwY-b\:EtN;lg
w:%vJw`q%o{%wДJ
数列 (2 14:21:46) 已知a1=2,an+1=(√2-1)(an +2)(n=1,2,3,……),则{an}的通项公式为an=
数列 (2 14:21:46)
已知a1=2,an+1=(√2-1)(an +2)(n=1,2,3,……),则{an}的通项公式为an=
数列 (2 14:21:46) 已知a1=2,an+1=(√2-1)(an +2)(n=1,2,3,……),则{an}的通项公式为an=
由 an+1=(√2-1)(an +2)
令:(an+1 + K)= (√2-1)(an + K)
解得 K = = -√2
所以(an+1 - √2)/(an - √2) = (√2-1)
又a1=2,所以 a1 -√2 = 2-√2=√2(√2-1)
所以 an = √2 (√2-1)^n + √2
an+1=(√2-1)(an +2)
an+2/an+1=√2+1=q
通项公式为an=a1q^(n-1)=2(√2-1)^(n-1)
这题目是不是出错了?就是等比数列啊
用待定系数法做
我算的也是等比数列