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
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![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](/uploads/image/z/11548600-16-0.jpg?t=%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E1%2Bn%2F2%E2%89%A41%2B1%2F2%2B1%2F3%2B...%2B1%2F%282%5En%29%E2%89%A41%2F2%2Bn%E5%BD%93n%3Dk%2B1%E6%97%B6%2C1%2B1%2F2%2B1%2F3%2B...%2B1%2F2%5Ek%2B1%2F%282%5Ek%2B1%29%2B...%2B1%2F2%5E%28k%2B1%29%3E%3D1%2Bk%2F2%2B1%2F%282%5Ek%2B1%29%2B...%2B1%2F2%5E%28k%2B1%29%3E1%2Bk%2F2%2B1%2F2%5E%28k%2B1%29%2B...%2B1%2F2%5E%28k%2B1%29%3E1%2Bk%2F2%2B%5B2%5E%28k%2B1%29-2%5Ek%5D%2F2%5E%28k%2B1%29%3D1%2B%28k%2B1%29%2F21%2B1%2F2%2B1%2F3%2B...%2B1%2F2%5Ek%2B1%2F%282%5Ek%2B1%29%2B...%2B1)
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 “ = ”
即
>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……