数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=21)当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立.(2)假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10那么 1/(k+2)+1/(k+3) +...+1/3(k+1)=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1)

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数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=21)当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立.(2)假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10那么 1/(k+2)+1/(k+3) +...+1/3(k+1)=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1)
数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=2
1)当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立.
(2)假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10
那么 1/(k+2)+1/(k+3) +...+1/3(k+1)
=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)
>9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)
=9/10
即n=k+1时命题也成立,
从而 原不等式对n∈N,且n>1成立.
第二步中为什么是
>9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)
不应该是
>9/10 +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)的么

数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=21)当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立.(2)假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10那么 1/(k+2)+1/(k+3) +...+1/3(k+1)=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1)
1)当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立.
(2)假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10
那么 1/(k+2)+1/(k+3) +...+1/3(k+1)
=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)
>9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)
=9/10
即n=k+1时命题也成立,
从而 原不等式对n∈N,且n>1成立.

写成这样是为了化简9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)=9/10+3/(3k+3)-1/(1+k)=9/10-1/(k+1)-1/(k+1)=9/101/(3k+3) +1/(3k+3)+1/(3k+3)=1/(3k+1) +1/(3k+2)+1/(3k+3) ??不等于,是小于 [1/(k+1) +1/(k+2)+...+1/3k] +1/(3k...

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写成这样是为了化简9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)=9/10+3/(3k+3)-1/(1+k)=9/10-1/(k+1)-1/(k+1)=9/10

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