已知数列{an}满足a1=2,an+1=2an/an +2则数列{an}的通项是

Question

如题所述

Answer 1

已知数列{an}满足dua1=2,a(n+1)=2an/(an
+2)则zhi数列{an}的通项是解：∵a(n+1)=2an/(an
+2)
dao∴内1/a（n+1）=（an+2）/2an=（1/2）+1/an
1/a（n+1）-1/an=1/2
令：bn=1/an
则b（n+1）=1/a（n+1）
b（n+1）-bn=1/2
b1=1/a1=1/2
∴bn=b1+（n-1）/2=（1/2）+[（n-1）/2]=n/2=1/an
an=1/bn=2/n
已知数列{an}满足a1=2,(n-1)an=n*a（n-1）(n>=2)则数列{an}的通容项是
解：∵(n-1)an=n*a（n-1）
∴an/a(n-1)=n/(n-1)
n≥2a2/a1=2/1a3/a2=3/2a4/a3=4/3
a(n-1)/a(n-2)=(n-1)/(n-2)
an/a(n-1)=n/(n-1)
以上式子两边相乘:

Answer 2

解：∵数列{a[n]}满足a[n
1]=(a[n]
2)/(a[n]
1)
采用不动点法，设：x=(x
2)/(x
1)
x^2=2
解得不动点是：x=±√2
∴(a[n
1]-√2)/(a[n
1]
√2)
={(a[n]
2)/(a[n]
1)-√2}/{(a[n]
2)/(a[n]
1)
√2}
={(a[n]
2)-√2(a[n]
1)}/{(a[n]
2)
√2(a[n]
1)}
={(1-√2)a[n]-(√2-2)}/{(1
√2)a[n]
(√2
2)}
={(1-√2)(a[n]-√2)}/{(1
√2)(a[n]
√2)}
={(1-√2)/(1
√2)}{(a[n]-√2)/(a[n]
√2)}
=(2√2-3){(a[n]-√2)/(a[n]
√2)}
∵a[1]=1
∴(a[1]-√2)/(a[1]
√2)=2√2-3
∴{(a[n]-√2)/(a[n]
√2)}是首项和公比均为2√2-3的等差数列
即：(a[n]-√2)/(a[n]
√2)=(2√2-3)(2√2-3)^(n-1)=(2√2-3)^n
a[n]-√2=a[n](2√2-3)^n
√2(2√2-3)^n
a[n][1-(2√2-3)^n]=√2[1
(2√2-3)^n]
∴{a[n]}的通项公式：a[n]=√2[1
(2√2-3)^n]/[1-(2√2-3)^n]