Tentukan jumlah deret aritmatika 1+2+3...+500 Tolong di jawab yaa

1+2+3...+500
n = 500
b = 2 - 1 = 1

Sn = ½n [2xU1 + (n - 1) b]
S500 = ½ x 500 [2 x 1 + (500 - 1) x 1]
= 250 (2 + 499)
= 250 x 501
= 125250