Question

Fie b1=2 in progresie geometrica. Determinati formula termenului general bn pentru bn+1=3bn.

Answer 1

[tex]\it b_1 = 2 \\\;\\ b_{n+1} = 3b_n \Longrightarrow b_{1+1}=3b_1\Longrightarrow b_2 = 3\cdot b_1 = 3\cdot2=6 \\\;\\ q = \dfrac{b_2}{b_1}=\dfrac{6}{2}=3[/tex]

$\it b_n = b_1\cdot q^{n-1} \Longrightarrow b_n = 2\cdot 3^{n-1}$