解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=128$$ 和公比：$$r=1.9140625$$ 插入几何级数的公式：

$$a_1=128$$

$$a_2=a_1·r^{n-1}=128\cdot {1.9140625}^{2-1}=128\cdot {1.9140625}^1=128\cdot 1.9140625=245$$

$$a_3=a_1·r^{n-1}=128\cdot {1.9140625}^{3-1}=128\cdot {1.9140625}^2=128\cdot 3.66363525390625=468.9453125$$

$$a_4=a_1·r^{n-1}=128\cdot {1.9140625}^{4-1}=128\cdot {1.9140625}^3=128\cdot 7.012426853179932=897.5906372070312$$

$$a_5=a_1·r^{n-1}=128\cdot {1.9140625}^{5-1}=128\cdot {1.9140625}^4=128\cdot 13.422223273664713=1718.0445790290833$$

$$a_6=a_1·r^{n-1}=128\cdot {1.9140625}^{6-1}=128\cdot {1.9140625}^5=128\cdot 25.690974234748865=3288.4447020478547$$

$$a_7=a_1·r^{n-1}=128\cdot {1.9140625}^{7-1}=128\cdot {1.9140625}^6=128\cdot 49.174130371199=6294.288687513472$$

$$a_8=a_1·r^{n-1}=128\cdot {1.9140625}^{8-1}=128\cdot {1.9140625}^7=128\cdot 94.12235891362309=12047.661940943755$$

$$a_9=a_1·r^{n-1}=128\cdot {1.9140625}^{9-1}=128\cdot {1.9140625}^8=128\cdot 180.15607760810667=23059.977933837654$$

$$a_{10}=a_1·r^{n-1}=128\cdot {1.9140625}^{10-1}=128\cdot {1.9140625}^9=128\cdot 344.8299922967667=44138.23901398614$$