解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=12,028$$ 和公比：$$r=1.0618556701030928$$ 插入几何级数的公式：

$$a_1=12028$$

$$a_2=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{2-1}=12028\cdot {1.0618556701030928}^1=12028\cdot 1.0618556701030928=12772$$

$$a_3=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{3-1}=12028\cdot {1.0618556701030928}^2=12028\cdot 1.1275374641300882=13562.0206185567$$

$$a_4=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{4-1}=12028\cdot {1.0618556701030928}^3=12028\cdot 1.1972820495401968=14400.908491869486$$

$$a_5=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{5-1}=12028\cdot {1.0618556701030928}^4=12028\cdot 1.27134073301691=15291.686336727393$$

$$a_6=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{6-1}=12028\cdot {1.0618556701030928}^5=12028\cdot 1.3499803659870282=16237.563842091975$$

$$a_7=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{7-1}=12028\cdot {1.0618556701030928}^6=12028\cdot 1.4334843061511742=17241.949234386324$$

$$a_8=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{8-1}=12028\cdot {1.0618556701030928}^7=12028\cdot 1.522153438490422=18308.461558162795$$

$$a_9=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{9-1}=12028\cdot {1.0618556701030928}^8=12028\cdot 1.616307259427974=19440.94371639967$$

$$a_{10}=a_1·r^{n-1}=12028\cdot {1.0618556701030928}^{10-1}=12028\cdot {1.0618556701030928}^9=12028\cdot 1.7162850280523847=20643.476317414083$$