解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=52$$ 和公比：$$r=1.2884615384615385$$ 插入几何级数的公式：

$$a_1=52$$

$$a_2=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{2-1}=52\cdot {1.2884615384615385}^1=52\cdot 1.2884615384615385=67$$

$$a_3=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{3-1}=52\cdot {1.2884615384615385}^2=52\cdot 1.6601331360946747=86.32692307692308$$

$$a_4=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{4-1}=52\cdot {1.2884615384615385}^3=52\cdot 2.1390176945835235=111.22892011834323$$

$$a_5=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{5-1}=52\cdot {1.2884615384615385}^4=52\cdot 2.75604202955954=143.31418553709608$$

$$a_6=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{6-1}=52\cdot {1.2884615384615385}^5=52\cdot 3.551054153470946=184.6548159804892$$

$$a_7=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{7-1}=52\cdot {1.2884615384615385}^6=52\cdot 4.575396697741412=237.92062828255342$$

$$a_8=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{8-1}=52\cdot {1.2884615384615385}^7=52\cdot 5.895222668243742=306.55157874867456$$

$$a_9=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{9-1}=52\cdot {1.2884615384615385}^8=52\cdot 7.595767668698668=394.97991877233073$$

$$a_{10}=a_1·r^{n-1}=52\cdot {1.2884615384615385}^{10-1}=52\cdot {1.2884615384615385}^9=52\cdot 9.7868544962079=508.9164338028108$$