f(k+1)有2^(k+1)-2项,f(k)有2^k-2项
因此f(k+1)-f(k)有[2^(k+1)-2]-[2^k-2]=2^k项,即
f(k+1)-f(k)=1/2^k+1/(2^k+1)+.+1/(2^(k+1)-2)+1/(2^(k+1)-1)