解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=1,125$$ 和公比：$$r=1.2$$ 插入几何级数的公式：

$$a_1=1125$$

$$a_2=a_1·r^{n-1}=1125\cdot {1.2}^{2-1}=1125\cdot {1.2}^1=1125\cdot 1.2=1350$$

$$a_3=a_1·r^{n-1}=1125\cdot {1.2}^{3-1}=1125\cdot {1.2}^2=1125\cdot 1.44=1620$$

$$a_4=a_1·r^{n-1}=1125\cdot {1.2}^{4-1}=1125\cdot {1.2}^3=1125\cdot 1.7279999999999998=1943.9999999999998$$

$$a_5=a_1·r^{n-1}=1125\cdot {1.2}^{5-1}=1125\cdot {1.2}^4=1125\cdot 2.0736=2332.7999999999997$$

$$a_6=a_1·r^{n-1}=1125\cdot {1.2}^{6-1}=1125\cdot {1.2}^5=1125\cdot 2.4883199999999994=2799.359999999999$$

$$a_7=a_1·r^{n-1}=1125\cdot {1.2}^{7-1}=1125\cdot {1.2}^6=1125\cdot 2.9859839999999993=3359.231999999999$$

$$a_8=a_1·r^{n-1}=1125\cdot {1.2}^{8-1}=1125\cdot {1.2}^7=1125\cdot 3.583180799999999=4031.078399999999$$

$$a_9=a_1·r^{n-1}=1125\cdot {1.2}^{9-1}=1125\cdot {1.2}^8=1125\cdot 4.2998169599999985=4837.294079999998$$

$$a_{10}=a_1·r^{n-1}=1125\cdot {1.2}^{10-1}=1125\cdot {1.2}^9=1125\cdot 5.1597803519999985=5804.752895999998$$