Răspuns:
Explicație pas cu pas:
b₃ = 12
b₃-b₂ = 6
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b₃ = b₁·q² = 12 => b₁ = 12/q²
b₂ = b₁·q
b₁·q²-b₁·q = 6 <=> b₁·q·(q-1) = 6 =>
q²-q = 6/b₁ = 6/(12/q²) = q²/2 =>
q²-2q = 0 <=> q(q-2) = 0 => q₁ = 0 ; q₂ = 2
b₁ = 12/4 = 3
b₂ = 3·2 = 6
b₃ = 6·2 = 12
b₂₀ = b₁·q¹⁹ => b₂₀ = 3·2¹⁹
S₃₀ = b₁+b₂+....+b₃₀ = b₁·(1+q+q²+....+q²⁹)
S₃₀ = 3·(1+2+2²+......+2²⁹) = 3·a =>
2·a = 2+2²+2³+....+2²⁹+2³⁰ = a-1+2³⁰ => a = 2³⁰-1
S₃₀ = 3·(2³⁰-1)
