Deret geometri dari 3+2+4/3+8/9+... Jumlah 8 suku pertama??

a = 3
r = 2/3
Sn = a(1 - rⁿ)/(1 - r)
S8 = 3(1 - (2/3)^8) / (1 - 2/3)
= 3(1 - (256/6561)) / (1/3)
= (3×3)(6561/6561 - 256/6561)
= 9(6305/6561)
= 6305/729
= 8(473/729)