解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=10$$ 和公比：$$r=0.9$$ 插入几何级数的公式：

$$a_1=10$$

$$a_2=a_1·r^{n-1}=10\cdot {0.9}^{2-1}=10\cdot {0.9}^1=10\cdot 0.9=9$$

$$a_3=a_1·r^{n-1}=10\cdot {0.9}^{3-1}=10\cdot {0.9}^2=10\cdot 0.81=8.100000000000001$$

$$a_4=a_1·r^{n-1}=10\cdot {0.9}^{4-1}=10\cdot {0.9}^3=10\cdot 0.7290000000000001=7.290000000000001$$

$$a_5=a_1·r^{n-1}=10\cdot {0.9}^{5-1}=10\cdot {0.9}^4=10\cdot 0.6561=6.561$$

$$a_6=a_1·r^{n-1}=10\cdot {0.9}^{6-1}=10\cdot {0.9}^5=10\cdot 0.5904900000000001=5.9049000000000005$$

$$a_7=a_1·r^{n-1}=10\cdot {0.9}^{7-1}=10\cdot {0.9}^6=10\cdot 0.531441=5.3144100000000005$$

$$a_8=a_1·r^{n-1}=10\cdot {0.9}^{8-1}=10\cdot {0.9}^7=10\cdot 0.4782969000000001=4.7829690000000005$$

$$a_9=a_1·r^{n-1}=10\cdot {0.9}^{9-1}=10\cdot {0.9}^8=10\cdot 0.4304672100000001=4.304672100000001$$

$$a_{10}=a_1·r^{n-1}=10\cdot {0.9}^{10-1}=10\cdot {0.9}^9=10\cdot 0.3874204890000001=3.874204890000001$$