Difference between revisions of "2020 CIME I Problems/Problem 7"

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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
 
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Notice that the product of first 2n+1 odd numbers = (2n+2)! / ((2^n+2)*(n+1)!) ; substituting this for f(n) gives the answer
 
==See also==
 
==See also==
 
{{CIME box|year=2020|n=I|num-b=6|num-a=8}}
 
{{CIME box|year=2020|n=I|num-b=6|num-a=8}}

Revision as of 00:37, 19 December 2020

Problem 7

For every positive integer $n$, define \[f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.\] Suppose that the sum $f(1)+f(2)+\cdots+f(2020)$ can be expressed as $\frac{p}{q}$ for relatively prime integers $p$ and $q$. Find the remainder when $p$ is divided by $1000$.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Notice that the product of first 2n+1 odd numbers = (2n+2)! / ((2^n+2)*(n+1)!) ; substituting this for f(n) gives the answer

See also

2020 CIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

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