用数学归纳法证明1+n/2≤1+1/2+1/3+...+1/(2^n)≤1/2+n当n=k+1时,1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)>=1+k/2+1/(2^k+1)+...+1/2^(k+1)>1+k/2+1/2^(k+1)+...+1/2^(k+1)>1+k/2+[2^(k+1)-2^k]/2^(k+1)=1+(k+1)/21+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/29 17:10:11

用数学归纳法证明1+n/2≤1+1/2+1/3+...+1/(2^n)≤1/2+n当n=k+1时,1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)>=1+k/2+1/(2^k+1)+...+1/2^(k+1)>1+k/2+1/2^(k+1)+...+1/2^(k+1)>1+k/2+[2^(k+1)-2^k]/2^(k+1)=1+(k+1)/21+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1
用数学归纳法证明1+n/2≤1+1/2+1/3+...+1/(2^n)≤1/2+n
当n=k+1时,
1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)
>=1+k/2+1/(2^k+1)+...+1/2^(k+1)
>1+k/2+1/2^(k+1)+...+1/2^(k+1)
>1+k/2+[2^(k+1)-2^k]/2^(k+1)=1+(k+1)/2
1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)

用数学归纳法证明1+n/2≤1+1/2+1/3+...+1/(2^n)≤1/2+n当n=k+1时,1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)>=1+k/2+1/(2^k+1)+...+1/2^(k+1)>1+k/2+1/2^(k+1)+...+1/2^(k+1)>1+k/2+[2^(k+1)-2^k]/2^(k+1)=1+(k+1)/21+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1
>=1+k/2+1/(2^k+1)+...+1/2^(k+1)
>1+k/2+1/2^(k+1)+...+1/2^(k+1) 到这行你还是懂的,对吧?
但是你还得明白一共有多少个1/2^(k+1)
一共有2^(k+1)-2^k个的1/2^(k+1)
接下去这行的,你的解答里有个小错误
“> ” 应该改成 “ = ”

>1+k/2+1/2^(k+1)+...+1/2^(k+1)
=1+k/2+[2^(k+1)-2^k]/2^(k+1)

“这一次的变化 是把2^(k+1)-2^k个的1/2^(k+1)和写成了” 这一句就是对你问题的直接解答!

[2^(k+1)-2^k]/2^(k+1)=1-1/2=1/2

还不清楚的话
hi我或者追问……