猜想:g(n)=n
即f(1)+f(2)+...f(n-1)=n[f(n)-1]
n=2时,左边=f(1)=1,右边=2*[f(2)-1]=1,左边=右边
假设n=k时,f(1)+f(2)+……+f(k-1)=k[f(k)-1]
当n=k+1时,左边=f(1)+f(2)+……+f(k-1)+f(k)
=k[f(k)-1]+f(k)
=(k+1)f(k)-k
=(k+1)[f(k+1)-1/(k+1)]-k
=(k+1)f(k+1)-1-k
=(k+1)[f(k+1)-1]
即n=k+1时成立
综上,f(1)+f(2)+...f(n-1)=n[f(n)-1]
故g(n)=n