Question

f(n)=1/2+1/3+1/4 ...+1/(2^n-1) ,则f(k+1)-f(k)=

Answer 1

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)