Question

已知数列{an}中，a1=0，an+1=an+2n-1(n属于N*)，求数列{an}的通项公式

求详细步骤，谢谢

Answer 1

详细步骤:a1=O和an+1=an+2n-1=>a2=a1+2*1-1=0+2-1=1=(2-1)^2=>a3=a2+2*2-1=1+3=4=(3-1)^2=>a4=a3+2*3-1=4+6-1=9=(4-1)^2 a1=0//a2=1//a3=4//a4=9可得an=(n-1)^2{n属于正整数}

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]