jumlah n suku pertama deret geometri dinyatakan dengan sn = 2 pangkat n+1+2pangkat n - 3 berapa deret rasio itu

[tex]s _{n} = {2}^{n + 1} + {2}^{n - 3} \\ s _{1} = {2}^{1 + 1} + {2}^{1 - 3} = {2}^{2} + \frac{1}{ {2}^{2} } = 4 + \frac{1}{4} = \frac{17}{4} \\ s _{2} = {2}^{2 + 1} + {2}^{2 - 3} = {2}^{3} + \frac{1}{2} = 8 + \frac{1}{2} = \frac{17}{2} \\ u _{n} = s_{n} - s _{n - 1} \\ u _{2} = \frac{17}{2} - \frac{17}{4} = \frac{17}{4} \\ r = \frac{u _{2} }{u _{1} } = \frac{ \frac{17}{4} }{ \frac{17}{4} } = 1[/tex]