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Compute

 

2020n=1[1n21(n+1)2] 

 Jul 26, 2021
edited by Guest  Jul 26, 2021
 #1
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Never mind this equation seems exceedingly obvious unless there's some trick I'm missing. 

 

2020n=1[1n21(n+1)2]=1114+1419+19...120202+120202120212

 

Thus, terms cancel and..

 

1120212

 Jul 26, 2021

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