Question

Rezolvati ecuatia: [tex] x^{2} = 1+3+5+...+101

Answer 1

x²=1+3+5+....+101
Termenii sunt in progresie aritmetica 
a1=1
r=2
n=51
an=a1+(n-1)*r= 1+(51-1)*2= 1+100=101
an=101
S=(a1+an)*n /2= 102*51/2 = 2601
x²= => x=√2601= 51

Answer 2

1+3+5+...+ (2n-1) =
2n-1 = 101 
2n = 101 + 1 ⇒ 2n =102 ⇒ n= 102 :2 ⇒ n = 51
n$x^{2} = 51· 51 x^{2} = 51^{2} x= 51$n