用数学归纳法证明等式1+
1
2 +
1
3 +…+
1
2 n -1 <f(n)(n≥2,n∈N *)的过程中,
假设n=k时不等式成立,左边=1+
1
2 +
1
3 +…+
1
2 k-1 ,
则当n=k+1时,左边=1+
1
2 +
1
3 +…+
1
2 k-1 +
1
2 k-1 +1 +…+
1
2 (k+1)-1 ,
∴由n=k递推到n=k+1时不等式左边增加了:
1
2 k-1 +1 +…+
1
2 (k+1)-1 =
1
2 k-1 +1 +…+
1
2 k ,
故选:D.
用数学归纳法证明等式1+
1
2 +
1
3 +…+
1
2 n -1 <f(n)(n≥2,n∈N *)的过程中,
假设n=k时不等式成立,左边=1+
1
2 +
1
3 +…+
1
2 k-1 ,
则当n=k+1时,左边=1+
1
2 +
1
3 +…+
1
2 k-1 +
1
2 k-1 +1 +…+
1
2 (k+1)-1 ,
∴由n=k递推到n=k+1时不等式左边增加了:
1
2 k-1 +1 +…+
1
2 (k+1)-1 =
1
2 k-1 +1 +…+
1
2 k ,
故选:D.