Question

用数学归纳法证明f(n)=1+1/2+1/4+...+1/2^n(n属于N）的过程中,从n=k到n=k+1时,f(k+1

Answer 1

f(k+1)=1+1/2+1/4+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)

f(K)=1+1/2+1/4+...+1/2^k

比较一下可知,f(k+1)比f(k)多的项为从

1/(2^k+1),1/(2^k+2),到1/2^(k+1),共2^(k+1)-（2^k+1）+1=2*2^K -2^k=2^k