Question

calculati a) √1+3+5+...+47/1+3+5+...+23 radical din tot

Answer 1

1+3+5+...+47
1=1+0x2
3=1+1x2
5=1+2x2
....
47=1+23x2 ⇒ 1+3+5+..+47 = 1 x 24 +2x(0+1+2+3+...+23)=
=24 + 2 x 23 x 24 :2 = 24 + 23x24 = 24(1+23) = 24 x 24


1+3+5+...+23
1=1+0x2
3=1+1x2
5=1+2x2
....
23=1+11x2 ⇒ 1+3+5+..+23 = 1 x 12 +2x(0+1+2+3+...+11)=
=12 + 2 x 11 x 12 :2 =12 + 11x12 = 12(1+12) =12x12

√24x24/12x12= √2x2 = 2

Answer 2

1+3+5+...+47 = numarul termenilor la patrat  adica: =24²

1+3+5+...+23 = 12²

a) exercitiul devine: $\sqrt{\frac{24^2}{12^2}}=\sqrt2^2=2$