解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=398$$ 和公比：$$r=1.2462311557788945$$ 插入几何级数的公式：

$$a_1=398$$

$$a_2=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{2-1}=398\cdot {1.2462311557788945}^1=398\cdot 1.2462311557788945=496$$

$$a_3=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{3-1}=398\cdot {1.2462311557788945}^2=398\cdot 1.5530920936339994=618.1306532663317$$

$$a_4=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{4-1}=398\cdot {1.2462311557788945}^3=398\cdot 1.935511754880562=770.3336784424637$$

$$a_5=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{5-1}=398\cdot {1.2462311557788945}^4=398\cdot 2.412095051308439=960.0138304207588$$

$$a_6=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{6-1}=398\cdot {1.2462311557788945}^5=398\cdot 3.0060280036406684=1196.399145448986$$

$$a_7=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{7-1}=398\cdot {1.2462311557788945}^6=398\cdot 3.746205753280833=1490.9898898057716$$

$$a_8=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{8-1}=398\cdot {1.2462311557788945}^7=398\cdot 4.668638325696716=1858.1180536272932$$

$$a_9=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{9-1}=398\cdot {1.2462311557788945}^8=398\cdot 5.8182025365466625=2315.6446095455717$$

$$a_{10}=a_1·r^{n-1}=398\cdot {1.2462311557788945}^{10-1}=398\cdot {1.2462311557788945}^9=398\cdot 7.250825271676243=2885.8284581271446$$