A Large Harmonic Mean

Algebra Level 4

The harmonic mean of the following numbers:

1 × 2 × 3 × 4 × … × 2013 , 1 × 2 × 3 × … × 2012 × 2014 , 1 × 2 × … × 2011 × 2013 × 2014 , ⋮ 2 × 3 × 4 × … × 2014 1 \times 2 \times 3 \times 4 \times \ldots \times 2013, \\ 1 \times 2 \times 3 \times \ldots \times 2012 \times 2014, \\ 1 \times 2 \times \ldots \times 2011 \times 2013 \times 2014, \\ \vdots \\ 2 \times 3 \times 4 \times \ldots \times 2014

can be expressed in the form a b 2014 ! , \frac{a}{b} 2014!, where a a and b b are coprime integers. Find a + b a+b .


The answer is 2017.

Michael Ng by Michael Ng , 22, United Kingdom

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2 solutions

Michael Ng
Nov 8, 2014

Notice that the numbers are 2014 ! 2014 , 2014 ! 2013 , … , 2014 ! 1 \frac{2014!}{2014}, \frac{2014!}{2013}, \ldots , \frac{2014!}{1} . So the harmonic mean is:

2014 2014 + 2013 + … + 1 2014 ! = 2014 × 2014 ! 2014 × 2015 2 = 2 2015 2014 ! \frac{2014}{\frac{2014+2013+\ldots +1}{2014!}}=\frac{2014 \times 2014!}{\frac{2014\times 2015}{2}} = \frac{2}{2015} 2014! giving the answer 2017 \boxed{2017} as required.

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Tushar Rao
Nov 17, 2014

1) Used the formula for Harmonic mean to calculate a generic expression 2) Applied the formula for calculating nth partial sum of all natural numbers

a/b = n/[n(n+1)]/2. Calculate this for n = 2014 to get a/b = 2/2015. Hence a+b = 2017

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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