解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=42$$ 和公比：$$r=1.2380952380952381$$ 插入几何级数的公式：

$$a_1=42$$

$$a_2=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{2-1}=42\cdot {1.2380952380952381}^1=42\cdot 1.2380952380952381=52$$

$$a_3=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{3-1}=42\cdot {1.2380952380952381}^2=42\cdot 1.5328798185941044=64.38095238095238$$

$$a_4=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{4-1}=42\cdot {1.2380952380952381}^3=42\cdot 1.8978512039736533=79.70975056689343$$

$$a_5=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{5-1}=42\cdot {1.2380952380952381}^4=42\cdot 2.3497205382530946=98.68826260662998$$

$$a_6=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{6-1}=42\cdot {1.2380952380952381}^5=42\cdot 2.909177809265736=122.18546798916091$$

$$a_7=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{7-1}=42\cdot {1.2380952380952381}^6=42\cdot 3.601839192424245=151.27724608181828$$

$$a_8=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{8-1}=42\cdot {1.2380952380952381}^7=42\cdot 4.459419952525256=187.29563800606073$$

$$a_9=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{9-1}=42\cdot {1.2380952380952381}^8=42\cdot 5.521186607888412=231.8898375313133$$

$$a_{10}=a_1·r^{n-1}=42\cdot {1.2380952380952381}^{10-1}=42\cdot {1.2380952380952381}^9=42\cdot 6.8357548478618435=287.10170361019743$$