Question

已知数列｛An｝满足：An＋1＝An＋2，A1＝3，前n项和为Sn.求｛An｝的通项公式。

Answer 1

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]