Question

Rezolvati va rog :
1+3+5+...+99=?

Answer 1

Se observa ca nr sunt in progresie aritmetica.
a1=1
a2=3
r=a2-a1=3-1=2
an=a1+(n-1)r
99=1+2(n-1)
99=1+2n-2
99=2n-1
100=2n
n=50
S=(a1+an)*n/2
S=(a1+a50)*50/2
S=(1+99)*25
S=100*25
S=2500

Answer 2

1 + 3  + .. + 2n - 1 = n²

2n - 1 = 99 => 2n = 100 => n = 50

1 + 3 + 5 +...+ 99 = 50² = 2500