解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=1,700$$ 和公比：$$r=1.2$$ 插入几何级数的公式：

$$a_1=1700$$

$$a_2=a_1·r^{n-1}=1700\cdot {1.2}^{2-1}=1700\cdot {1.2}^1=1700\cdot 1.2=2040$$

$$a_3=a_1·r^{n-1}=1700\cdot {1.2}^{3-1}=1700\cdot {1.2}^2=1700\cdot 1.44=2448$$

$$a_4=a_1·r^{n-1}=1700\cdot {1.2}^{4-1}=1700\cdot {1.2}^3=1700\cdot 1.7279999999999998=2937.5999999999995$$

$$a_5=a_1·r^{n-1}=1700\cdot {1.2}^{5-1}=1700\cdot {1.2}^4=1700\cdot 2.0736=3525.12$$

$$a_6=a_1·r^{n-1}=1700\cdot {1.2}^{6-1}=1700\cdot {1.2}^5=1700\cdot 2.4883199999999994=4230.143999999999$$

$$a_7=a_1·r^{n-1}=1700\cdot {1.2}^{7-1}=1700\cdot {1.2}^6=1700\cdot 2.9859839999999993=5076.1727999999985$$

$$a_8=a_1·r^{n-1}=1700\cdot {1.2}^{8-1}=1700\cdot {1.2}^7=1700\cdot 3.583180799999999=6091.407359999998$$

$$a_9=a_1·r^{n-1}=1700\cdot {1.2}^{9-1}=1700\cdot {1.2}^8=1700\cdot 4.2998169599999985=7309.688831999997$$

$$a_{10}=a_1·r^{n-1}=1700\cdot {1.2}^{10-1}=1700\cdot {1.2}^9=1700\cdot 5.1597803519999985=8771.626598399998$$