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：$$r=1.2950819672131149$$ 插入几何级数的公式：

$$a_1=61$$

$$a_2=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{2-1}=61\cdot {1.2950819672131149}^1=61\cdot 1.2950819672131149=79$$

$$a_3=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{3-1}=61\cdot {1.2950819672131149}^2=61\cdot 1.6772373018005915=102.31147540983608$$

$$a_4=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{4-1}=61\cdot {1.2950819672131149}^3=61\cdot 2.1721597842991267=132.50174684224672$$

$$a_5=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{5-1}=61\cdot {1.2950819672131149}^4=61\cdot 2.8131249665513285=171.60062295963104$$

$$a_6=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{6-1}=61\cdot {1.2950819672131149}^5=61\cdot 3.6432274156976225=222.23687235755497$$

$$a_7=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{7-1}=61\cdot {1.2950819672131149}^6=61\cdot 4.718278128526429=287.81496584011217$$

$$a_8=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{8-1}=61\cdot {1.2950819672131149}^7=61\cdot 6.110556920550622=372.74397215358795$$

$$a_9=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{9-1}=61\cdot {1.2950819672131149}^8=61\cdot 7.913672077434413=482.73399672349916$$

$$a_{10}=a_1·r^{n-1}=61\cdot {1.2950819672131149}^{10-1}=61\cdot {1.2950819672131149}^9=61\cdot 10.248854001923258=625.1800941173187$$