解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=96$$ 和公比：$$r=1.125$$ 插入几何级数的公式：

$$a_1=96$$

$$a_2=a_1·r^{n-1}=96\cdot {1.125}^{2-1}=96\cdot {1.125}^1=96\cdot 1.125=108$$

$$a_3=a_1·r^{n-1}=96\cdot {1.125}^{3-1}=96\cdot {1.125}^2=96\cdot 1.265625=121.5$$

$$a_4=a_1·r^{n-1}=96\cdot {1.125}^{4-1}=96\cdot {1.125}^3=96\cdot 1.423828125=136.6875$$

$$a_5=a_1·r^{n-1}=96\cdot {1.125}^{5-1}=96\cdot {1.125}^4=96\cdot 1.601806640625=153.7734375$$

$$a_6=a_1·r^{n-1}=96\cdot {1.125}^{6-1}=96\cdot {1.125}^5=96\cdot 1.802032470703125=172.9951171875$$

$$a_7=a_1·r^{n-1}=96\cdot {1.125}^{7-1}=96\cdot {1.125}^6=96\cdot 2.0272865295410156=194.6195068359375$$

$$a_8=a_1·r^{n-1}=96\cdot {1.125}^{8-1}=96\cdot {1.125}^7=96\cdot 2.2806973457336426=218.9469451904297$$

$$a_9=a_1·r^{n-1}=96\cdot {1.125}^{9-1}=96\cdot {1.125}^8=96\cdot 2.565784513950348=246.3153133392334$$

$$a_{10}=a_1·r^{n-1}=96\cdot {1.125}^{10-1}=96\cdot {1.125}^9=96\cdot 2.8865075781941414=277.1047275066376$$