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的最大值,并证明结论
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![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的最大值,并证明结论](/uploads/image/z/657848-56-8.jpg?t=%E8%8B%A5%E4%B8%8D%E7%AD%89%E5%BC%8F1%2F%EF%BC%88n%2B1%EF%BC%89%2B1%2F%EF%BC%88n%2B2%29%2B%E2%80%A6%E2%80%A61%2F%EF%BC%883n%2B1%EF%BC%89%EF%BC%9Ea%2F24%E5%AF%B9%E4%B8%80%E5%88%87%E6%AD%A3%E6%95%B4%E6%95%B0n%E9%83%BD%E6%88%90%E7%AB%8B%2C%E6%B1%82%E6%AD%A3%E6%95%B4%E6%95%B0a%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%2C%E5%B9%B6%E8%AF%81%E6%98%8E%E7%BB%93%E8%AE%BA)
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的最大值,并证明结论
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的最大值,并证明结论
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的最大值,并证明结论
f(n)=1/(n+1) + 1/(n+2) +1/(n+3) +……+1/(3n+1)
f(n+1)=1/(n+2) + 1/(n+3) +1/(n+4) +……+1/[3(n+1)+1]
f(n+1)-f(n)=1/(n+1) - 1/(3n+2)-1/(3n+3)-1/(3n+4)>0
所以函数f(n)对于n为正整数时为单调增函数
所以原不等式等效于a/2425/24
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/[(n+1)+1]+1/[(n+1)+2]+1/[(n+1)+3]+...+1/[3(n+1)+1]
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{1/(3n+2)+1/(3n+3)+1/(3n+4)-1/(n+1)}
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{6(n+1)/[(3n+2)(3n+4)]-2/(3n+3)}
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+2/(3n+3)*{9(n+1)^2/[(3n+2)(3n+4)]-1}
>1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24.
结论成立.
这样可以么?