1. Oblicz sumy S₆ i S₇ ciągu geometrycznego (an) o ilorazie q.
a.
a₁ = 81, q= -⅓
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Sn = a1 * (1 - q^n) / (1 - q)
S6 = 81 * (1 - (-1/3)^6) / (1 + 1/3)
S6 = 81 * (1 - 1/729) / (4/3)
S6 = 81 * 728/729 * 3/4
S6 = 182/3
S7 = 81 * (1 - (-1/3)^7) / (1 + 1/3)
S6 = 81 * (1 + 1/2187) / (4/3)
S7 = 81 * 2188/2187 * 3/4
S7 = 547/9
S6 + S7 = 182/3 + 547/9 = 546/9 + 547/9 = 1093/9 = 121 i 4/9