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1+3+5+....+(2n-1)=n pătrat, n mai mare sau egal cu delta

Răspuns :

   
[tex]\displaystyle 1+3+5+...+(2n-1)=n^2 \\ \texttt{Calculam numarul de termeni din sir: } \\ \\ \frac{(2n-1)-1}{2}+1 = \frac{2n-2}{2}+1 = \frac{2(n-1)}{2}+1 =n-1+1 = n\,termeni \\ \\ 1+3+5+...+(2n-1)=n^2 \\ \\ \frac{n[(2n-1)+1]}{2}=n^2 \\ \\ \frac{n[2n-1+1]}{2}=n^2 \\ \\ \frac{n \times 2n}{2}=n^2 \\ \\ n \times n=n^2 \\ cctd[/tex]



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