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.
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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.
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.
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.