解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=400$$ 和公比：$$r=1.1025$$ 插入几何级数的公式：

$$a_1=400$$

$$a_2=a_1·r^{n-1}=400\cdot {1.1025}^{2-1}=400\cdot {1.1025}^1=400\cdot 1.1025=441$$

$$a_3=a_1·r^{n-1}=400\cdot {1.1025}^{3-1}=400\cdot {1.1025}^2=400\cdot 1.21550625=486.2025$$

$$a_4=a_1·r^{n-1}=400\cdot {1.1025}^{4-1}=400\cdot {1.1025}^3=400\cdot 1.3400956406250002=536.0382562500001$$

$$a_5=a_1·r^{n-1}=400\cdot {1.1025}^{5-1}=400\cdot {1.1025}^4=400\cdot 1.4774554437890628=590.9821775156252$$

$$a_6=a_1·r^{n-1}=400\cdot {1.1025}^{6-1}=400\cdot {1.1025}^5=400\cdot 1.6288946267774416=651.5578507109766$$

$$a_7=a_1·r^{n-1}=400\cdot {1.1025}^{7-1}=400\cdot {1.1025}^6=400\cdot 1.7958563260221294=718.3425304088518$$

$$a_8=a_1·r^{n-1}=400\cdot {1.1025}^{8-1}=400\cdot {1.1025}^7=400\cdot 1.9799315994393978=791.9726397757591$$

$$a_9=a_1·r^{n-1}=400\cdot {1.1025}^{9-1}=400\cdot {1.1025}^8=400\cdot 2.182874588381936=873.1498353527744$$

$$a_{10}=a_1·r^{n-1}=400\cdot {1.1025}^{10-1}=400\cdot {1.1025}^9=400\cdot 2.4066192336910848=962.6476934764339$$