Question

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Answer 1

Explicație pas cu pas:

√(1+3+5+...+2013)∈Q

√[(2-1)+(4-1)+(6-1)+...+(2014-1)]

=√[2(1+2+3+...+1007)-1007])

=√[2*1007(1007+1)/2-1007]

=√(1007*1008-1007)

=√(1007(1008-1)

=√1007²

1007∈Q

Bafta!

Answer 2

1+3+5+...+n = [(n+1)/2]² (formulă)

1+3+5+...+2013 = [(2013+1)/2]² =

= (2014/2)² = 1007²

=> √(1+3+5+...+2013) = √(1007²) =

= |1007| = 1007 ∈ Q