[tex]\textbf{Suma lui Gauss:}\\ 1+2+3+...+n = \dfrac{n\cdot (n+1)}{2}[/tex]
Metoda I:
1+3+5+...+(2·n-1) = n·n (formulă)
Deci:
1+3+5+...+19 =
= 1+3+5+...+(2·10-1)
= 10·10
= 100
[tex]\\[/tex]
Metoda II:
1+3+5+...+19 =
= (2·1-1)+(2·2-1)+(2·3-1)+...+(2·10-1)
= 2·1 + 2·2 + 2·3 + ... + 2·10 - 1 - 1 - 1 -...- 1 {de 10 ori}
= 2·(1+2+3+...+10) - 1·10
= 2·[10·(10+1)]/2 - 10
= 10·(10+1) - 10
= 10·11 - 10
= 100
[tex]\\[/tex]
Metoda III:
1+3+5+...+19 =
= 1+2+3+...+9+
+1+2+...+10
= (1+2+3+...+9) + (1+2+3+...+10)
= [9·(9+1)]/2 + [10·(10+1)]/2
= (9·10)/2 + (10·11)/2
= 9·5 + 5·11
= 100
[tex]\\[/tex]
Metoda IV:
[tex]\text{Formula:}\\1+3+5+...+n = \dfrac{(n+1)(n+1)}{4}\\\\\text{Aplicare:}\\ b)\quad 1+3+5+...+19 =\dfrac{(19+1)(19+1)}{4} = \dfrac{20\cdot 20}{4} = \\ \\ = 20\cdot 5 = 100[/tex]
