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注:一定要用完整的数学归纳法解题

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注:一定要用完整的数学归纳法解题
①求证:1/(n+1) +1/(n+2) +……+1/3n >5/6 (n≥2)
②求证:当n≥5时,2的n次方>n的平方
第2小题好象有点问题啊...
注:一定要用完整的数学归纳法解题
1.
当n=2时,
1/3+1/4+1/5+1/6=57/60=19/20>5/6
设当n=k时,不等式成立
1/(k+1)+1/(k+2)+……+1/3k>5/6
当n=k+1时
1/(k+1+1)+1/(k+1+2)+……+1/3k+1/(3k+1)+1/(3k+2)+1/(3k+3)
=1/(k+1)+1/(k+2)+……+1/3k+[1/(3k+1)+1/(3k+2)+1/(3k+3)-1/(k+1))]
1/(k+1)+1/(k+2)+……+1/3k>5/6
1/(3k+1)+1/(3k+2)+1/(3k+3)-1/(k+1)>0
(因为1/(k+1)=3*1/(3k+3))
所以当n=k+1时
1/(k+1+1)+1/(k+1+2)+……+1/3k+1/(3k+1)+1/(3k+2)+1/(3k+3)>5/6
2.
当n=5时
2^5=32>5^2
设当n=k时成立,2^k>k^2
当n=k+1时
2^(k+1)=2^k+2^k>2*k^2
(k+1)^2=k^2+2k+1(k+1)^2