Question

a1=1,an+1=2an²,求an的通项公式

Answer 1

用数学归纳法解.

a2=2a1²=2×1²=2=2^(2^0)

a3=2a2²=2×2^2=2^3=2^(2^0+2^1)

假设n=k (k∈N+且k≥1)时

ak=2^[2^0+2^1+...+2^(k-2)]=2^[(2^(k-1)-1)/(2-1)]=2^(2^(k-1)-1)

则当n=k+1时,

a(k+1)=2ak²=2×2^[2(2^(k-1)-1)]

=2^[2^(k)-2+1]

=2^[2^(k)-1]

同样成立.

综上,得an=2^(2^(ⁿ-1)-1)

^表示指数,2^(2^(ⁿ-1)-1)表示2的2(ⁿ-1)-1次方.