Question

(Bn)n>1 progresie geometrica, Sn=b1+b2+…+bn. S3=21; S6=189; Sn=?

Answer 1

Răspuns:

Explicație pas cu pas:

S₃=21; S₆=189

S₃ = b₁+b₂+b₃ = b₁+b₁q+b₁q² = b₁(1+q+q²)

S₆ = S₃+b₄+b₅+b₆ = 21+b₁q³+b₁q⁴+b₁q⁵ =

= 21+b₁q³(1+q+q²) = 189 =>

21+21·q² = 189 => q²+1 = 189:21 = 9 => q² = 8

q₁,₂ = ±2√2

b₁ = 21/(1+8±2√2) = 21/(9±2√2)

Sₙ = b₁·(qⁿ⁻¹-1)/(q-1) = [21/(9±2√2)]·(±2√2ⁿ⁻¹-1)/(±2√2-1)

Answer 2

$\it S_3=b_1+b_2+b_3 \Rightarrow 21=b_1+b_1q+b_1q^2 \Rightarrow b_1(1+q+q^2)=21\ \ \ \ \ (1)\\ \\ \\ S_6=S_3+b_4+b_5+b_6 \Rightarrow 189=21+b_4(1+q+q^2)\bigg|_{-21} \Rightarrow \\ \\ \\ \Rightarrow b_4(1+q+q^2)=168\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow \dfrac{b_4(1+q+q^2)}{b_1(1+q+q^2)}=\dfrac{168}{21} \Rightarrow \dfrac{b_4}{b_1}=8 \Rightarrow \dfrac{b_1q^3}{b_1}=2^3 \Rightarrow q=2\ \ \ (3)$

$\it (1),\ (3) \Rightarrow b_1(1+2+4)=21 \Rightarrow b_1\cdot7=21 \Rightarrow b_1=3\\ \\ \\ S_n=b_1\cdot\dfrac{q^n-1}{q-1}=3\cdot\dfrac{2^n-1}{2-1}=3(2^n-1)$