已知f(x)=ax-ln(-x)……证明:2^0/(2^0+1)+2^1/(2^1+1)+2^2/(2^2+1)+……+

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/17 20:46:21

已知f(x)=ax-ln(-x)……证明:2^0/(2^0+1)+2^1/(2^1+1)+2^2/(2^2+1)+……+2^(n-1)/(2^(n-1)+1)>ln((2^n+1)/2),n为正整数.

可以验证n=1,2时满足.假设当n=k时,2^0/(2^0+1)+2^1/(2^1+1)+2^2/(2^2+1)+……+2^(k-1)/(2^(k-1)+1)>ln((2^k+1)/2)
则当n=k+1时,2^0/(2^0+1)+2^1/(2^1+1)+2^2/(2^2+1)+……+2^(k-1)/(2^(k-1)+1) + 2^k/(2^k+1) > ln((2^k+1)/2) + 2^k/(2^k+1)
ln((2^(k+1) +1)/2)-ln((2^k+1)/2)=ln( ((2^(k+1) +1)/2) / ((2^k+1)/2) )
=ln(((2^(k+1) +1) / (2^k+1) )=ln(((2·(2^k+1-1) +1) / (2^k+1) )=ln(2 - 1 / (2^k+1) )现在需要证明 ln(2 - 1 / (2^k+1) ) - 2^k/(2^k+1)<0,即 ln(2 - 1 / (2^k+1) ) -1+ 1/(2^k+1)<0先证明当0<x<1/2时,ln(2 - x ) -1+ x <0令f(x)=ln(2 - x ) -1+ x ,则f'(x)=1/(x-2)+1>0.说明当0<x<1/2时,f(x)=ln(2 - x ) -1+ x是单调递增函数.而f(1/2)=ln(3/2 ) -1/2(3/2)²=9/4=2.25<e,→3/2<√e→ln(3/2 ) <1/2→f(1/2) <0.所以当0<x<1/2时,增函数f(x)=ln(2 - x ) -1+ x<0.
当k≥1时,0< 1 / (2^k+1)<1/2→f(1 / (2^k+1) )= ln(2 - 1 / (2^k+1) ) -1+ 1/(2^k+1)<0→ln(2 - 1 / (2^k+1) ) - 2^k/(2^k+1)<0,ln(2 - 1 / (2^k+1) ) < 2^k/(2^k+1)即 ln((2^(k+1) +1)/2) -ln((2^k+1)/2) < 2^k/(2^k+1)→ln((2^k+1)/2) + 2^k/(2^k+1) > ln((2^(k+1) +1)/2)
因此,当n=k+1时,有 2^0/(2^0+1)+2^1/(2^1+1)+2^2/(2^2+1)+……+2^(k-1)/(2^(k-1)+1) + 2^k/(2^k+1) > ln((2^(k+1) +1)/2).∴n为正整数时,2^0/(2^0+1)+2^1/(2^1+1)+2^2/(2^2+1)+……+2^(n-1)/(2^(n-1)+1)>ln((2^n+1)/2)成立.

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