分式求和问题1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)为什么1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)=1/(2^k+1)+1/(2^k+2)+…+1/[2^k+2^(k-1)]+1/[(2^k+2^(k-1)+1]+1/[(2^k+2^(k-1)+2]+…+1/2^(k+1)≥2^(k-1)/[2^k+2^(k-1)]+2^(k-1)/2^(k+1)=7/12如何由第二步

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分式求和问题1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)为什么1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)=1/(2^k+1)+1/(2^k+2)+…+1/[2^k+2^(k-1)]+1/[(2^k+2^(k-1)+1]+1/[(2^k+2^(k-1)+2]+…+1/2^(k+1)≥2^(k-1)/[2^k+2^(k-1)]+2^(k-1)/2^(k+1)=7/12如何由第二步
分式求和问题1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)
为什么1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)
=1/(2^k+1)+1/(2^k+2)+…+1/[2^k+2^(k-1)]+1/[(2^k+2^(k-1)+1]+1/[(2^k+2^(k-1)+2]+…+1/2^(k+1)
≥2^(k-1)/[2^k+2^(k-1)]+2^(k-1)/2^(k+1)=7/12
如何由第二步到第三步的?

分式求和问题1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)为什么1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)=1/(2^k+1)+1/(2^k+2)+…+1/[2^k+2^(k-1)]+1/[(2^k+2^(k-1)+1]+1/[(2^k+2^(k-1)+2]+…+1/2^(k+1)≥2^(k-1)/[2^k+2^(k-1)]+2^(k-1)/2^(k+1)=7/12如何由第二步
第二步中,前半部分分母全部变为2^k+2^(k-1),后半部分分母全部变为2^(k+1),然后分别相加.
分母增大,则结果变小

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