解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=61$$ 和公比：$$r=1.0491803278688525$$ 插入几何级数的公式：

$$a_1=61$$

$$a_2=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{2-1}=61\cdot {1.0491803278688525}^1=61\cdot 1.0491803278688525=64$$

$$a_3=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{3-1}=61\cdot {1.0491803278688525}^2=61\cdot 1.1007793603869929=67.14754098360656$$

$$a_4=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{4-1}=61\cdot {1.0491803278688525}^3=61\cdot 1.154916050242091=70.44987906476754$$

$$a_5=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{5-1}=61\cdot {1.0491803278688525}^4=61\cdot 1.211715200253997=73.91462721549382$$

$$a_6=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{6-1}=61\cdot {1.0491803278688525}^5=61\cdot 1.271307751086161=77.54977281625582$$

$$a_7=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{7-1}=61\cdot {1.0491803278688525}^6=61\cdot 1.3338310831067919=81.3636960695143$$

$$a_8=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{8-1}=61\cdot {1.0491803278688525}^7=61\cdot 1.3994293330956507=85.3651893188347$$

$$a_9=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{9-1}=61\cdot {1.0491803278688525}^8=61\cdot 1.4682537265265843=89.56347731812164$$

$$a_{10}=a_1·r^{n-1}=61\cdot {1.0491803278688525}^{10-1}=61\cdot {1.0491803278688525}^9=61\cdot 1.5404629261918261=93.96823849770139$$