解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=57$$ 和公比：$$r=1.3508771929824561$$ 插入几何级数的公式：

$$a_1=57$$

$$a_2=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{2-1}=57\cdot {1.3508771929824561}^1=57\cdot 1.3508771929824561=77$$

$$a_3=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{3-1}=57\cdot {1.3508771929824561}^2=57\cdot 1.82486919052016=104.01754385964912$$

$$a_4=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{4-1}=57\cdot {1.3508771929824561}^3=57\cdot 2.465174169650041=140.51492767005232$$

$$a_5=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{5-1}=57\cdot {1.3508771929824561}^4=57\cdot 3.330147562509704=189.81841106305313$$

$$a_6=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{6-1}=57\cdot {1.3508771929824561}^5=57\cdot 4.498620391460477=256.4213623132472$$

$$a_7=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{7-1}=57\cdot {1.3508771929824561}^6=57\cdot 6.077083686709767=346.39377014245673$$

$$a_8=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{8-1}=57\cdot {1.3508771929824561}^7=57\cdot 8.209393752221967=467.93544387665213$$

$$a_9=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{9-1}=57\cdot {1.3508771929824561}^8=57\cdot 11.089882788089323=632.1233189210914$$

$$a_{10}=a_1·r^{n-1}=57\cdot {1.3508771929824561}^{10-1}=57\cdot {1.3508771929824561}^9=57\cdot 14.98106973127856=853.9209746828778$$