解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=57$$ 和公比：$$r=1.9473684210526316$$ 插入几何级数的公式：

$$a_1=57$$

$$a_2=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{2-1}=57\cdot {1.9473684210526316}^1=57\cdot 1.9473684210526316=111$$

$$a_3=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{3-1}=57\cdot {1.9473684210526316}^2=57\cdot 3.79224376731302=216.15789473684214$$

$$a_4=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{4-1}=57\cdot {1.9473684210526316}^3=57\cdot 7.384895757399039=420.9390581717452$$

$$a_5=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{5-1}=57\cdot {1.9473684210526316}^4=57\cdot 14.381112790724444=819.7234290712934$$

$$a_6=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{6-1}=57\cdot {1.9473684210526316}^5=57\cdot 28.005324908252867=1596.3035197704135$$

$$a_7=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{7-1}=57\cdot {1.9473684210526316}^6=57\cdot 54.53668534765032=3108.5910648160684$$

$$a_8=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{8-1}=57\cdot {1.9473684210526316}^7=57\cdot 106.20301883489799=6053.5720735891855$$

$$a_9=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{9-1}=57\cdot {1.9473684210526316}^8=57\cdot 206.8164050995382=11788.535090673677$$

$$a_{10}=a_1·r^{n-1}=57\cdot {1.9473684210526316}^{10-1}=57\cdot {1.9473684210526316}^9=57\cdot 402.7477362464691=22956.62096604874$$