Question

2•(1+2+3+........+2019):2020

Answer 1

2•(1+2+3+........+2019):2020=

pentru (1+2+3+....+2019) folosim suma lui Gauss:
s= n(n+1)/2
s= 2019(2019+1)/2
s= 2019•2020/2
s= 4078380/2
s= 2039190

2•(1+2+3+........+2019):2020=
2• 2039190: 2020=
4078380:2020= 2019

Answer 2

2•(1+2+3+........+2019):2020=

=2× [2019×(2019+1):2]:2020=
=2×(2019×2020:2):2020=
=2×(2019×1010):2020=
=2×2039190:2020=
=2×2020=
=4040

dar deoarece le vom calcula in ordinea in care vin =>
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2•(1+2+3+........+2019):2020=

=2× [2019×(2019+1):2]:2020=
=2×(2019×2020:2):2020=
=2×(2019×1010):2020=
=2×2039190:2020=
=4078380:2020=
=2019
ultima e corectă