解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=27$$ 和公比：$$r=1.4814814814814814$$ 插入几何级数的公式：

$$a_1=27$$

$$a_2=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{2-1}=27\cdot {1.4814814814814814}^1=27\cdot 1.4814814814814814=40$$

$$a_3=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{3-1}=27\cdot {1.4814814814814814}^2=27\cdot 2.194787379972565=59.25925925925925$$

$$a_4=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{4-1}=27\cdot {1.4814814814814814}^3=27\cdot 3.2515368592186147=87.7914951989026$$

$$a_5=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{5-1}=27\cdot {1.4814814814814814}^4=27\cdot 4.817091643286836=130.06147436874457$$

$$a_6=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{6-1}=27\cdot {1.4814814814814814}^5=27\cdot 7.136432064128646=192.68366573147344$$

$$a_7=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{7-1}=27\cdot {1.4814814814814814}^6=27\cdot 10.572491946857252=285.4572825651458$$

$$a_8=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{8-1}=27\cdot {1.4814814814814814}^7=27\cdot 15.662951032381113=422.89967787429003$$

$$a_9=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{9-1}=27\cdot {1.4814814814814814}^8=27\cdot 23.20437189982387=626.5180412952444$$

$$a_{10}=a_1·r^{n-1}=27\cdot {1.4814814814814814}^{10-1}=27\cdot {1.4814814814814814}^9=27\cdot 34.376847258998325=928.1748759929548$$