Lời giải:Ta có: \({a_{n + 1}} - k = q\left( {{a_n} - k} \right) \Leftrightarrow k - kq = 3 \Leftrightarrow k = \frac{3}{{1 - q}}\)
Đặt \({v_n} = {a_n} - k \Rightarrow {v_{n + 1}} = q.{v_n} = {q^2}.{v_{n - 1}} = ... = {q^n}{v_1}\)
Khi đó \({v_n} = {q^{n - 1}}.{v_1} = {q^{n - 1}}.\left( {{a_1} - k} \right) = {q^{n - 1}}.\left( {5 - \frac{3}{{1 - q}}} \right)\)
Vậy \({a_n} = {v_n} + k = {q^{n - 1}}.\left( {5 - \frac{3}{{1 - q}}} \right) + k = {q^{n - 1}}.\left( {5 - \frac{3}{{1 - q}}} \right) + \frac{3}{{1 - q}} = 5{q^{n - 1}} + 3\frac{{1 - {q^{n - 1}}}}{{1 - q}}\)
Do dó: \(\alpha = 5;\beta = 3 \Rightarrow \alpha + 2\beta = 5 + 2.3 = 11\)
