Question

)Fie progresia geometrica in care b1=6 si b2=3. Sa se afle suma primilor 10 termeni. *​

Answer 1

Răspuns:

3(4 - 1/2⁸)

Explicație pas cu pas:

b₁ = 6 ; b₂ = 3  ; progresie geometrica ; S₁₀ = ?

b₂ = b₁q => q = b₂:b₁ = 3/6 => q = 1/2

b₃ = 3· 1/2 = 3/2

b₄ = 3/2 ·1/2 = 3/4

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b₁₀ = b₄·q⁶ = 3/4 · (1/2⁶) = 3/4 · 1/64 = 3/256

S₁₀ = 6+3+3/2+3/4+...............+3/256 = 9 + 3·(1/2 + 1/2²+1/2³+....+1/2⁸)

S₁₀ = 9 + 3·a

a = 1/2 + 1/2²+1/2³+....+1/2⁸   I·2 =>

2a = 1+1/2+1/2²+....+1/2⁷ = a + 1 - 1/2⁸ =>

a = 1-1/2⁸ => S¹⁰ = 9+3·(1-1/2⁸) = 9+3-3/2⁸ = 12 - 3/2⁸ =

= 3(4 - 1/2⁸) = 11,98828125

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