解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=32$$ 和公比：$$r=2.1875$$ 插入几何级数的公式：

$$a_1=32$$

$$a_2=a_1·r^{n-1}=32\cdot {2.1875}^{2-1}=32\cdot {2.1875}^1=32\cdot 2.1875=70$$

$$a_3=a_1·r^{n-1}=32\cdot {2.1875}^{3-1}=32\cdot {2.1875}^2=32\cdot 4.78515625=153.125$$

$$a_4=a_1·r^{n-1}=32\cdot {2.1875}^{4-1}=32\cdot {2.1875}^3=32\cdot 10.467529296875=334.9609375$$

$$a_5=a_1·r^{n-1}=32\cdot {2.1875}^{5-1}=32\cdot {2.1875}^4=32\cdot 22.897720336914062=732.72705078125$$

$$a_6=a_1·r^{n-1}=32\cdot {2.1875}^{6-1}=32\cdot {2.1875}^5=32\cdot 50.08876323699951=1602.8404235839844$$

$$a_7=a_1·r^{n-1}=32\cdot {2.1875}^{7-1}=32\cdot {2.1875}^6=32\cdot 109.56916958093643=3506.213426589966$$

$$a_8=a_1·r^{n-1}=32\cdot {2.1875}^{8-1}=32\cdot {2.1875}^7=32\cdot 239.68255845829844=7669.84187066555$$

$$a_9=a_1·r^{n-1}=32\cdot {2.1875}^{9-1}=32\cdot {2.1875}^8=32\cdot 524.3055966275278=16777.77909208089$$

$$a_{10}=a_1·r^{n-1}=32\cdot {2.1875}^{10-1}=32\cdot {2.1875}^9=32\cdot 1146.9184926227172=36701.39176392695$$