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若 $a<\frac{1}{\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{...

Mar 8, 2024

若 a<\frac{1}{\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}}<a+1 ，则自然数

a=

Solution by Steps

step 1

Calculate the sum of the reciprocals:

\frac{1}{2011} + \frac{1}{2012} + \frac{1}{2013} + \frac{1}{2014} + \frac{1}{2015}

step 2

Find the reciprocal of the sum to get the middle expression of the inequality:

\frac{1}{\left(\frac{1}{2011} + \frac{1}{2012} + \frac{1}{2013} + \frac{1}{2014} + \frac{1}{2015}\right)}

step 3

Solve the inequality a < \frac{1}{\left(\frac{1}{2011} + \frac{1}{2012} + \frac{1}{2013} + \frac{1}{2014} + \frac{1}{2015}\right)} < a + 1 for

a

step 4

The solution to the inequality is given by the interval \frac{16485733536834043}{41050153620137} < a < \frac{16526783690454180}{41050153620137}

step 5

Determine the natural number

a

that satisfies the inequality

Answer

a = 402

Key Concept

Solving inequalities involving fractions and natural numbers

Explanation

The natural number

a

is the integer part of the lower bound of the solution interval for the inequality, which is the greatest integer less than the upper bound of the interval.