当n=1时 左边=1/2+1/3+1/4=13/12>1，成立
假设n=k时成立 即1/(k+1)+1/(k+2)+1/(k+3)...+1/(3k+1)>1
当n=k+1时 即要证明 1/(k+2)+1/(k+3)+...+1/(3k+1)+1/(3k+2)+1/(3k+3)+1/(3k+4)>1
式子里比n=k的式子的左边多了 1/(3k+2)+1/(3k+3)+1/(3k+4)，少了1/(k+1)
所以 只需要证明 1/(3k+2)+1/(3k+3)+1/(3k+4)>1/(k+1)即可
而 1/(3k+2)+1/(3k+3)+1/(3k+4)
=1/(3k+3)+(3k+4+3k+2)/(3k+2)(3k+4)
=1/(3k+3)+(6k+6)/(9k²+18k+8)
＞1/(3k+3)+(6k+6)/(9k²+18k+9)
=1/(3k+3)+(6k+6)/(3k+3)²
=1/(3k+3)+2/(3k+3)
=1/(k+1)
所以 n=k+1时也成立
所以对一切正整数n，均有1/(n+1)+1/(n+2)+...+1/(3n+1)>1
求证1\/n+1+1\/n+2+...+1\/3n+1>1(n属于正整数)
=1\/(3k+3)+(6k+6)\/(3k+3)²=1\/(3k+3)+2\/(3k+3)=1\/(k+1)所以 n=k+1时也成立 所以对一切正整数n,均有1\/(n+1)+1\/(n+2)+...+1\/(3n+1)>1
求证1\/n+1+1\/n+2+...+1\/3n+1>1(n属于正整数)
=1\/(3k+3)+(6k+6)\/(3k+3)²=1\/(3k+3)+2\/(3k+3)=1\/(k+1)所以 n=k+1时也成立 所以对一切正整数n，均有1\/(n+1)+1\/(n+2)+...+1\/(3n+1)>1
...+1\/(n+2)+1\/(n+3)+……+1\/(3n+1)>1(N属于正整数)
f(n+1)-f(n)=1\/(3n+2)+1\/(3n+3)+1\/(3n+4)-1\/(n+1)=2\/(3n+2)(3n+3)(3n+4)>0 f(n)递增 所以f(n)最小值为f(1)=13\/12
若不等式1\/n+1+1\/n+2……+1\/3n+1>a\/24对一切正整数n都成立,求正整数a...
设f(n)=1\/(n+1)+1\/(n+2)+…+1\/(3n+1)则f(n+1)-f(n)=1\/(3n+2)+1\/(3n+3)+1\/(3n+4)-1\/(n+1)>0 故f(n)单调增加，要使f(n)>a\/24对一切正整数n都成立,只要 f(1)=1\/2+1\/3+1\/4>a\/24 即a<26 故正整数a的最大值为25 ...
求证:1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/(3n+1)>25\/24(n是正整数)_百度知 ...
证明：当k=1时 1\/2+1\/3+1\/4=13\/12=26\/24>25\/24 结论成立.假设k=n时结论成立,即 1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/(3n+1)>25\/24 当k=n+1时 由于 9(n+1)^2=9n^2+18n+9>9n^2+18n+8=(3n+2)(3n+4)即 9(n+1)^2\/[(3n+2)(3n+4)]-1>0 左侧为 1...
求证1\/(n+1)+1\/(n+2)+.+1\/(3n+1)>1 [n属于N*]
1\/(n+1)+1\/(3n+1)>2\/(2n+1)1\/(n+2)+1\/(3n)>2\/(2n+1).1\/(2n)+1\/(2n+2)>2\/(2n+1)1\/(2n+1)=1\/(2n+1)1\/(n+1)+1\/(n+2)+.+1\/(3n+1)>(2n+1)\/(2n+1)=1
求证:n属于正整数,1\/(n+1)+1\/(n+2)~~+1\/2n>=2n\/3n+1
用数学归纳法，当n=1时不等式成立。若结论对n成立，则有1\/(n+2)+...+1\/2n+1\/2n+1+1\/(2n+2)>=2n\/(3n+1)+1\/(2n+1)+1\/(2n+2)-1\/(n+1)=2n\/(3n+1)+1\/(2n+1)-1\/(2n+2)=2n\/(3n+1)+1\/(2n+1)(2n+2)>(2n+2)\/(3n+4)，最后一个不等式是因为(倒推)1\/(...
若不等式1\/n+1+1\/n+2……+1\/3n+1>a\/24对一切正整数n都成立,求正整数a...
令sn=1\/n+1+1\/n+2……+1\/3n+1 则s(n+1)=1\/(n+2)+1\/(n+3)+...1\/(3n+4)s(n+1)-sn=1\/(3n+2)+ 1\/(3n+3)+ 1\/(3n+4)- 1\/(n+1)=1\/(3n+2)+ 1\/(3n+3)+ 1\/(3n+4)- 1\/3(n+1)- 1\/3(n+1)- 1\/3(n+1)=[1\/(3n+2)-1\/(3n+3)]-[1\/(3n+3...
如何证明:1\/n+1+1\/n+2+L+1\/3n>5\/6(n>=2 n属于正整数)
1\/(n+1)+1\/(n+2)+...+1\/3n=1\/3+1\/4+1\/5+1\/6+1\/(n+5)+...+1\/3n=19\/20+...+1\/3n≥19\/20＞5\/6
求证:1\/(n+1)+1\/(n+2)+...+1\/3n>5\/6(n大于等于2,且是整数!)
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