1+3+5+...+39=n^2. Atunci n=?

Răspuns:

1+3+5+...+39=

= 1+2+3+4+...+39 -2-4-6-...-38= ( am adunat 2,4,6,8 si toate nr pare pana la 38, apoi le-am scazut)

= 39*40/2 - (2+4+6+...+38)

= 780 - 2*(1+2+3+...+19)

= 780- 2* 19*20/2

= 780-19*20= 780- 380= 400

deci 1+3+5+...+39=400

si 1+3+5+...+39= n²

=> n²=400 => n²= 20² => n= ±20

Răspuns:

1 + 2 + 5 +...+ 39 = n^2

= 2 + 4 + 6 +... + 40 - 20

= 2 × ( 1 + 2 + 3 + ... + 20 ) - 20

= 2× [ 20 + ( 20 + 1 ) ] : 2 - 20

= 2 × ( 20 × 21 ) : 2 - 20

= 2 × 420 : 2 - 20

= 420 - 20

= 400

400 = 20^2 => n = 20