Question

已知数列｛an}满足an+1=an+2^n，且a1=1，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]