解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=4$$ 和公比：$$r=8.25$$ 插入几何级数的公式：

$$a_1=4$$

$$a_2=a_1·r^{n-1}=4\cdot {8.25}^{2-1}=4\cdot {8.25}^1=4\cdot 8.25=33$$

$$a_3=a_1·r^{n-1}=4\cdot {8.25}^{3-1}=4\cdot {8.25}^2=4\cdot 68.0625=272.25$$

$$a_4=a_1·r^{n-1}=4\cdot {8.25}^{4-1}=4\cdot {8.25}^3=4\cdot 561.515625=2246.0625$$

$$a_5=a_1·r^{n-1}=4\cdot {8.25}^{5-1}=4\cdot {8.25}^4=4\cdot 4632.50390625=18530.015625$$

$$a_6=a_1·r^{n-1}=4\cdot {8.25}^{6-1}=4\cdot {8.25}^5=4\cdot 38218.1572265625=152872.62890625$$

$$a_7=a_1·r^{n-1}=4\cdot {8.25}^{7-1}=4\cdot {8.25}^6=4\cdot 315299.7971191406=1261199.1884765625$$

$$a_8=a_1·r^{n-1}=4\cdot {8.25}^{8-1}=4\cdot {8.25}^7=4\cdot 2601223.32623291=10404893.30493164$$

$$a_9=a_1·r^{n-1}=4\cdot {8.25}^{9-1}=4\cdot {8.25}^8=4\cdot 21460092.44142151=85840369.76568604$$

$$a_{10}=a_1·r^{n-1}=4\cdot {8.25}^{10-1}=4\cdot {8.25}^9=4\cdot 177045762.64172745=708183050.5669098$$