解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=38$$ 和公比：$$r=1.105263157894737$$ 插入几何级数的公式：

$$a_1=38$$

$$a_2=a_1·r^{n-1}=38\cdot {1.105263157894737}^{2-1}=38\cdot {1.105263157894737}^1=38\cdot 1.105263157894737=42.00000000000001$$

$$a_3=a_1·r^{n-1}=38\cdot {1.105263157894737}^{3-1}=38\cdot {1.105263157894737}^2=38\cdot 1.2216066481994463=46.42105263157896$$

$$a_4=a_1·r^{n-1}=38\cdot {1.105263157894737}^{4-1}=38\cdot {1.105263157894737}^3=38\cdot 1.350196821694125=51.30747922437675$$

$$a_5=a_1·r^{n-1}=38\cdot {1.105263157894737}^{5-1}=38\cdot {1.105263157894737}^4=38\cdot 1.4923228029250855=56.70826651115325$$

$$a_6=a_1·r^{n-1}=38\cdot {1.105263157894737}^{6-1}=38\cdot {1.105263157894737}^5=38\cdot 1.6494094137593052=62.6775577228536$$

$$a_7=a_1·r^{n-1}=38\cdot {1.105263157894737}^{7-1}=38\cdot {1.105263157894737}^6=38\cdot 1.8230314573129165=69.27519537789082$$

$$a_8=a_1·r^{n-1}=38\cdot {1.105263157894737}^{8-1}=38\cdot {1.105263157894737}^7=38\cdot 2.0149295054511183=76.5673212071425$$

$$a_9=a_1·r^{n-1}=38\cdot {1.105263157894737}^{9-1}=38\cdot {1.105263157894737}^8=38\cdot 2.227027348130184=84.62703922894698$$

$$a_{10}=a_1·r^{n-1}=38\cdot {1.105263157894737}^{10-1}=38\cdot {1.105263157894737}^9=38\cdot 2.4614512795123087=93.53514862146773$$