解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=32$$ 和公比：$$r=1.40625$$ 插入几何级数的公式：

$$a_1=32$$

$$a_2=a_1·r^{n-1}=32\cdot {1.40625}^{2-1}=32\cdot {1.40625}^1=32\cdot 1.40625=45$$

$$a_3=a_1·r^{n-1}=32\cdot {1.40625}^{3-1}=32\cdot {1.40625}^2=32\cdot 1.9775390625=63.28125$$

$$a_4=a_1·r^{n-1}=32\cdot {1.40625}^{4-1}=32\cdot {1.40625}^3=32\cdot 2.780914306640625=88.9892578125$$

$$a_5=a_1·r^{n-1}=32\cdot {1.40625}^{5-1}=32\cdot {1.40625}^4=32\cdot 3.910660743713379=125.14114379882812$$

$$a_6=a_1·r^{n-1}=32\cdot {1.40625}^{6-1}=32\cdot {1.40625}^5=32\cdot 5.499366670846939=175.97973346710205$$

$$a_7=a_1·r^{n-1}=32\cdot {1.40625}^{7-1}=32\cdot {1.40625}^6=32\cdot 7.733484380878508=247.47150018811226$$

$$a_8=a_1·r^{n-1}=32\cdot {1.40625}^{8-1}=32\cdot {1.40625}^7=32\cdot 10.875212410610402=348.00679713953286$$

$$a_9=a_1·r^{n-1}=32\cdot {1.40625}^{9-1}=32\cdot {1.40625}^8=32\cdot 15.293267452420878=489.3845584774681$$

$$a_{10}=a_1·r^{n-1}=32\cdot {1.40625}^{10-1}=32\cdot {1.40625}^9=32\cdot 21.50615735496686=688.1970353589395$$