證：
f'(x)=x²-1
a(n+1)=(an +1)²-1=an²+2an=an(an+2)
a1≥1 a1-(2-1)=a1-1≥0,不等式成立.
假設(shè)當(dāng)n=k(k∈N+)時(shí),不等式,即ak≥2^k -1,則當(dāng)n=k+1時(shí),
a(k+1)=ak(ak+2)≥(2^k -1)(2^k -1+2)=2^(2k)-1
k≥1 2k-k≥1 2k≥k+1
a(k+1)≥2^(k+1)-1
不等式同樣成立.
綜上,得an≥2ⁿ-1.