解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=12,500$$ 和公比：$$r=1.22$$ 插入几何级数的公式：

$$a_1=12500$$

$$a_2=a_1·r^{n-1}=12500\cdot {1.22}^{2-1}=12500\cdot {1.22}^1=12500\cdot 1.22=15250$$

$$a_3=a_1·r^{n-1}=12500\cdot {1.22}^{3-1}=12500\cdot {1.22}^2=12500\cdot 1.4884=18605$$

$$a_4=a_1·r^{n-1}=12500\cdot {1.22}^{4-1}=12500\cdot {1.22}^3=12500\cdot 1.815848=22698.1$$

$$a_5=a_1·r^{n-1}=12500\cdot {1.22}^{5-1}=12500\cdot {1.22}^4=12500\cdot 2.2153345599999996=27691.681999999997$$

$$a_6=a_1·r^{n-1}=12500\cdot {1.22}^{6-1}=12500\cdot {1.22}^5=12500\cdot 2.7027081631999996=33783.85204$$

$$a_7=a_1·r^{n-1}=12500\cdot {1.22}^{7-1}=12500\cdot {1.22}^6=12500\cdot 3.2973039591039996=41216.299488799996$$

$$a_8=a_1·r^{n-1}=12500\cdot {1.22}^{8-1}=12500\cdot {1.22}^7=12500\cdot 4.022710830106879=50283.88537633599$$

$$a_9=a_1·r^{n-1}=12500\cdot {1.22}^{9-1}=12500\cdot {1.22}^8=12500\cdot 4.907707212730393=61346.34015912991$$

$$a_{10}=a_1·r^{n-1}=12500\cdot {1.22}^{10-1}=12500\cdot {1.22}^9=12500\cdot 5.987402799531079=74842.53499413849$$