证明:数学归纳法(1)当n=1时,an=1+√2=a+b√2 (a、b∈Z)对应得到a=1,为奇数,b=1(2)假设n=k (k∈N*)时,ak=(1+√2)^k=a(k)+b(k)√2,(a、b∈Z),a(k)为奇数[a(k)的k为角标](3)n=k+1时,a(k+1)=(1+√2)^(k+1)=a(k+1) + b(k+1)√2得到 (1+√2)^k * (1+√2)=[a(k)+b(k)√2] * (1+√2)=a(k+1) + b(k+1)√2即a(k)+b(k)*2+[a(k)+b(k)]√2=a(k+1) + b(k+1)√2对应得到 a(k+1)=a(k)+b(k)*2,∵a(k)为奇数,b(k)为整数,∴a(k+1)为奇数(4)综上:a是奇数