解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=61$$ 和公比：$$r=1.0163934426229508$$ 插入几何级数的公式：

$$a_1=61$$

$$a_2=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{2-1}=61\cdot {1.0163934426229508}^1=61\cdot 1.0163934426229508=62$$

$$a_3=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{3-1}=61\cdot {1.0163934426229508}^2=61\cdot 1.0330556302069336=63.01639344262295$$

$$a_4=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{4-1}=61\cdot {1.0163934426229508}^3=61\cdot 1.0499909684070474=64.04944907282989$$

$$a_5=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{5-1}=61\cdot {1.0163934426229508}^4=61\cdot 1.0672039351022449=65.09944004123693$$

$$a_6=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{6-1}=61\cdot {1.0163934426229508}^5=61\cdot 1.084699081579331=66.16664397633919$$

$$a_7=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{7-1}=61\cdot {1.0163934426229508}^6=61\cdot 1.102481033736369=67.25134305791852$$

$$a_8=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{8-1}=61\cdot {1.0163934426229508}^7=61\cdot 1.1205544933058178=68.35382409165489$$

$$a_9=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{9-1}=61\cdot {1.0163934426229508}^8=61\cdot 1.1389242390977166=69.4743785849607$$

$$a_{10}=a_1·r^{n-1}=61\cdot {1.0163934426229508}^{10-1}=61\cdot {1.0163934426229508}^9=61\cdot 1.1575951282632528=70.61330282405842$$