Jumlah 5 suku pertama dari deret geometri deret 1 + 2/3 + 4/9 + adalah

U1 = a = 1
U2 = a×r = 1×2/3 >>> r = 2/3

S5 = {1 × [1-(2/3)^5]} / (1 - 2/3)

S5 = [1 × (1 - 32/243)] / [(3-2)/3]

S5 = {1 × [(243-32)/243]} / (1/3)

S5 = (1 × 211/243) / (1/3)

S5 = (211/243) / (1/3)

S5 = (211×3)/(243×1)

S5 = 633/243

S5 = 2 147/243 atau 2 49/81

coba dicek lg jawabannya, maaf kl ada yg salah