求证:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整数)
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/10 13:34:57
求证:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整数)
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/[(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.
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/[(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.
求证:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整数)
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
求证不等式(3^n-4^n)大于等于4^(n-1)其中n属于正整数
当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)..
证明(n-2)n(n+1)(n+3)+9(n为正整数)是完全平方数
n为正整数,f(n)为正整数,f(n)为n的增函数.f[f(n)]=2n+1,求证:4/3
求证:(n+2002)(n+2003)(n+2004)(n+2005)+1是一个完全平方数(n为正整数)
若n为正整数,求1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/
1\n(n+3)+1\(n+3)(n+6)+1\(n+6)(n+9)=1\2 n+18 n为正整数,求n的值
求证1/(n+1)+1/(n+2)+.+1/(3n+1)>1 [n属于N*]
对于任意正整数n,求证:ln(1/2+1/n)>1/n^2-2/n-1
n为正整数,则 n(n+1)(n+2)(n+3)+1的值一定是某个数的平方