解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=99$$ 和公比：$$r=1.0303030303030303$$ 插入几何级数的公式：

$$a_1=99$$

$$a_2=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{2-1}=99\cdot {1.0303030303030303}^1=99\cdot 1.0303030303030303=102$$

$$a_3=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{3-1}=99\cdot {1.0303030303030303}^2=99\cdot 1.061524334251607=105.0909090909091$$

$$a_4=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{4-1}=99\cdot {1.0303030303030303}^3=99\cdot 1.0936917383198375=108.27548209366391$$

$$a_5=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{5-1}=99\cdot {1.0303030303030303}^4=99\cdot 1.1268339122083173=111.55655730862341$$

$$a_6=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{6-1}=99\cdot {1.0303030303030303}^5=99\cdot 1.160980394396448=114.93705904524836$$

$$a_7=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{7-1}=99\cdot {1.0303030303030303}^6=99\cdot 1.1961616184690678=118.42000022843771$$

$$a_8=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{8-1}=99\cdot {1.0303030303030303}^7=99\cdot 1.2324089402408576=122.0084850838449$$

$$a_9=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{9-1}=99\cdot {1.0303030303030303}^8=99\cdot 1.2697546657027017=125.70571190456747$$

$$a_{10}=a_1·r^{n-1}=99\cdot {1.0303030303030303}^{10-1}=99\cdot {1.0303030303030303}^9=99\cdot 1.3082320798149047=129.51497590167557$$