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 1+3+5+...+(2n-1)? Cum se rezolva pas cu pas?

Răspuns :

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Se cunoaste formula  [tex]1+2+...+n=\dfrac{n(n+1)}{2}.[/tex]

Prin urmare, avem  [tex]1+2+...+2n=\dfrac{2n(2n+1)}{2}=n(2n+1)[/tex] .

Observam ca :

[tex]1+3+...+(2n-1)=[1+2+...+2n]-[2+4+...+2n]\\ \\ =[1+2+...+2n]-[2(1+2+...+n)]\\ \\ =n(2n+1)-2\dfrac{n(n+1)}{2}\\ \\ =n(2n+1)-n(n+1) \\ \\ =n(2n+1-n-1)\\ \\ =n^2[/tex]

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