解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=308$$ 和公比：$$r=1.2857142857142858$$ 插入几何级数的公式：

$$a_1=308$$

$$a_2=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{2-1}=308\cdot {1.2857142857142858}^1=308\cdot 1.2857142857142858=396.00000000000006$$

$$a_3=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{3-1}=308\cdot {1.2857142857142858}^2=308\cdot 1.6530612244897962=509.1428571428572$$

$$a_4=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{4-1}=308\cdot {1.2857142857142858}^3=308\cdot 2.125364431486881=654.6122448979594$$

$$a_5=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{5-1}=308\cdot {1.2857142857142858}^4=308\cdot 2.732611411911704=841.6443148688049$$

$$a_6=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{6-1}=308\cdot {1.2857142857142858}^5=308\cdot 3.513357529600763=1082.114119117035$$

$$a_7=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{7-1}=308\cdot {1.2857142857142858}^6=308\cdot 4.517173966629553=1391.2895817219023$$

$$a_8=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{8-1}=308\cdot {1.2857142857142858}^7=308\cdot 5.8077950999522825=1788.800890785303$$

$$a_9=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{9-1}=308\cdot {1.2857142857142858}^8=308\cdot 7.467165128510078=2299.886859581104$$

$$a_{10}=a_1·r^{n-1}=308\cdot {1.2857142857142858}^{10-1}=308\cdot {1.2857142857142858}^9=308\cdot 9.600640879512959=2956.9973908899915$$