解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=380$$ 和公比：$$r=1.105263157894737$$ 插入几何级数的公式：

$$a_1=380$$

$$a_2=a_1·r^{n-1}=380\cdot {1.105263157894737}^{2-1}=380\cdot {1.105263157894737}^1=380\cdot 1.105263157894737=420.00000000000006$$

$$a_3=a_1·r^{n-1}=380\cdot {1.105263157894737}^{3-1}=380\cdot {1.105263157894737}^2=380\cdot 1.2216066481994463=464.2105263157896$$

$$a_4=a_1·r^{n-1}=380\cdot {1.105263157894737}^{4-1}=380\cdot {1.105263157894737}^3=380\cdot 1.350196821694125=513.0747922437674$$

$$a_5=a_1·r^{n-1}=380\cdot {1.105263157894737}^{5-1}=380\cdot {1.105263157894737}^4=380\cdot 1.4923228029250855=567.0826651115325$$

$$a_6=a_1·r^{n-1}=380\cdot {1.105263157894737}^{6-1}=380\cdot {1.105263157894737}^5=380\cdot 1.6494094137593052=626.775577228536$$

$$a_7=a_1·r^{n-1}=380\cdot {1.105263157894737}^{7-1}=380\cdot {1.105263157894737}^6=380\cdot 1.8230314573129165=692.7519537789083$$

$$a_8=a_1·r^{n-1}=380\cdot {1.105263157894737}^{8-1}=380\cdot {1.105263157894737}^7=380\cdot 2.0149295054511183=765.6732120714249$$

$$a_9=a_1·r^{n-1}=380\cdot {1.105263157894737}^{9-1}=380\cdot {1.105263157894737}^8=380\cdot 2.227027348130184=846.2703922894699$$

$$a_{10}=a_1·r^{n-1}=380\cdot {1.105263157894737}^{10-1}=380\cdot {1.105263157894737}^9=380\cdot 2.4614512795123087=935.3514862146773$$