解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=5,000$$ 和公比：$$r=1.2$$ 插入几何级数的公式：

$$a_1=5000$$

$$a_2=a_1·r^{n-1}=5000\cdot {1.2}^{2-1}=5000\cdot {1.2}^1=5000\cdot 1.2=6000$$

$$a_3=a_1·r^{n-1}=5000\cdot {1.2}^{3-1}=5000\cdot {1.2}^2=5000\cdot 1.44=7200$$

$$a_4=a_1·r^{n-1}=5000\cdot {1.2}^{4-1}=5000\cdot {1.2}^3=5000\cdot 1.7279999999999998=8639.999999999998$$

$$a_5=a_1·r^{n-1}=5000\cdot {1.2}^{5-1}=5000\cdot {1.2}^4=5000\cdot 2.0736=10368$$

$$a_6=a_1·r^{n-1}=5000\cdot {1.2}^{6-1}=5000\cdot {1.2}^5=5000\cdot 2.4883199999999994=12441.599999999997$$

$$a_7=a_1·r^{n-1}=5000\cdot {1.2}^{7-1}=5000\cdot {1.2}^6=5000\cdot 2.9859839999999993=14929.919999999996$$

$$a_8=a_1·r^{n-1}=5000\cdot {1.2}^{8-1}=5000\cdot {1.2}^7=5000\cdot 3.583180799999999=17915.903999999995$$

$$a_9=a_1·r^{n-1}=5000\cdot {1.2}^{9-1}=5000\cdot {1.2}^8=5000\cdot 4.2998169599999985=21499.084799999993$$

$$a_{10}=a_1·r^{n-1}=5000\cdot {1.2}^{10-1}=5000\cdot {1.2}^9=5000\cdot 5.1597803519999985=25798.901759999993$$