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：$$r=1.2727272727272727$$ 插入几何级数的公式：

$$a_1=99$$

$$a_2=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{2-1}=99\cdot {1.2727272727272727}^1=99\cdot 1.2727272727272727=126$$

$$a_3=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{3-1}=99\cdot {1.2727272727272727}^2=99\cdot 1.6198347107438016=160.36363636363635$$

$$a_4=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{4-1}=99\cdot {1.2727272727272727}^3=99\cdot 2.061607813673929=204.09917355371897$$

$$a_5=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{5-1}=99\cdot {1.2727272727272727}^4=99\cdot 2.6238644901304555=259.7625845229151$$

$$a_6=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{6-1}=99\cdot {1.2727272727272727}^5=99\cdot 3.3394638965296704=330.60692575643736$$

$$a_7=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{7-1}=99\cdot {1.2727272727272727}^6=99\cdot 4.250226777401399=420.7724509627385$$

$$a_8=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{8-1}=99\cdot {1.2727272727272727}^7=99\cdot 5.409379534874508=535.5285739525763$$

$$a_9=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{9-1}=99\cdot {1.2727272727272727}^8=99\cdot 6.884664862567555=681.581821394188$$

$$a_{10}=a_1·r^{n-1}=99\cdot {1.2727272727272727}^{10-1}=99\cdot {1.2727272727272727}^9=99\cdot 8.762300734176888=867.467772683512$$