Question

高一数学，已知a1=2,an+1=(根号2-1)(an+2)求an ？求过程，在线等，谢谢了

Answer 1

1.
a(n+1)=(√2-1)a(n+2)
a(n+2)/a(n+1)=1/(√2-1)=√2+1
an=a1(√2+1)^(n-1)
=2(√2+1)^(n-1)

2.
a(n+1)=(√2-1)(an+2)
x=(√2-1)(x+2)
x=√2
a(n+1)-√2=(√2-1)(an+2)-√2
=(√2-1)an+2(√2-1)-√2
=(√2-1)an-√2(√2-1)
=(√2-1)(an-√2)
设bn=an-√2,b1=a1-√2=2-√2
b(n+1)=(√2-1)bn
bn=b1(√2-1)^(n-1)
=(2-√2)(√2-1)^(n-1)
=√2(√2-1)^n
an=bn+√2
=√2+√2(√2-1)^n

追问

不好意思，可能我题目没说清楚，是:已知a1=2 ， a(n+1)=(√2-1)，求(an)

Answer 2

题目意思？！！！
是a(n+1) 还是 (an)+1

追问

a(n+1)

追答

后边是 (an)+2 还是a(n+2)

追问

(an)+2

追答

a(n+1)=(√2-1)(an+2)
x=(√2-1)(x+2)
x=√2
a(n+1)-√2=(√2-1)(an+2)-√2
=(√2-1)an+2(√2-1)-√2
=(√2-1)an-√2(√2-1)
=(√2-1)(an-√2)
设bn=an-√2,b1=a1-√2=2-√2
b(n+1)=(√2-1)bn
bn=b1(√2-1)^(n-1)
=(2-√2)(√2-1)^(n-1)
=√2(√2-1)^n
an=bn+√2
=√2+√2(√2-1)^n