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：$$r=1.2307692307692308$$ 插入几何级数的公式：

$$a_1=52$$

$$a_2=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{2-1}=52\cdot {1.2307692307692308}^1=52\cdot 1.2307692307692308=64$$

$$a_3=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{3-1}=52\cdot {1.2307692307692308}^2=52\cdot 1.5147928994082842=78.76923076923077$$

$$a_4=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{4-1}=52\cdot {1.2307692307692308}^3=52\cdot 1.8643604915794267=96.94674556213019$$

$$a_5=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{5-1}=52\cdot {1.2307692307692308}^4=52\cdot 2.294597528097756=119.31907146108333$$

$$a_6=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{6-1}=52\cdot {1.2307692307692308}^5=52\cdot 2.824120034581854=146.8542417982564$$

$$a_7=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{7-1}=52\cdot {1.2307692307692308}^6=52\cdot 3.475840042562282=180.74368221323866$$

$$a_8=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{8-1}=52\cdot {1.2307692307692308}^7=52\cdot 4.2779569754612705=222.45376272398607$$

$$a_9=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{9-1}=52\cdot {1.2307692307692308}^8=52\cdot 5.265177815952333=273.7892464295213$$

$$a_{10}=a_1·r^{n-1}=52\cdot {1.2307692307692308}^{10-1}=52\cdot {1.2307692307692308}^9=52\cdot 6.4802188504028715=336.9713802209493$$