Question

已知数列an满足a1=1 an=a1+1/2a2+1/3a3+•••1/n-1•an-1(n>1)

求an通项公式 谢谢

Answer 1

a1=1
a2=a1=1
a3=a1+a2/2=1+1/2=3/2
a4=a1+a2/2+a3/3=1+1/2+1/2=2
a5=a1+a2/2+a3/3+a4/4=1+1/2+1/2+1/2=5/2
猜想:当n>1时,有an=n/2
证明:(1)当n=2时,有a2=2/2=1成立.
(2)设当n=k时,有ak=k/2
则当n=k+1时,有ak+1=a1+a2/2+a3/3+......+ak-1/(k-1)+ak/k
而a1+a2/2+a3/3+......+ak-1/(k-1)=ak
∴ak+1=ak+ak/k=ak*(1+1/k)=k/2*(k+1)/k=(k+1)/2
即当n=k+1时,有ak+1=(k+1)/2成立
由(1)(2)可知,对於n>1时,有an=n/2成立.
∴an=1(n=1),an=n/2(n>1)