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)](/uploads/image/z/3274647-15-7.jpg?t=%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E1%2Fn%2B1%2B1%2Fn%2B2%2B1%2Fn%2B3%2B...%2B1%2F3n%3E9%2F10+n%3E%3D21%EF%BC%89%E5%BD%93n%3D2%E6%97%B6%2C%E5%B7%A6%3D1%2F3+%2B1%2F4%2B1%2F5%2B1%2F6%3D57%2F60%3E54%2F60%3D9%2F10%2C%E6%88%90%E7%AB%8B%EF%BC%8E%EF%BC%882%EF%BC%89%E5%81%87%E8%AE%BEn%3Dk%E6%97%B6%2C%E6%9C%891%2F%28k%2B1%29+%2B1%2F%28k%2B2%29+%2B...%2B1%2F3k+%3E9%2F10%E9%82%A3%E4%B9%88%E3%80%801%2F%28k%2B2%29%2B1%2F%28k%2B3%29+%2B...%2B1%2F3%28k%2B1%29%3D%5B1%2F%28k%2B1%29+%2B1%2F%28k%2B2%29%2B...%2B1%2F3k%5D+%2B1%2F%283k%2B1%29)
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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