解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=308$$ 和公比：$$r=1.7954545454545454$$ 插入几何级数的公式：

$$a_1=308$$

$$a_2=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{2-1}=308\cdot {1.7954545454545454}^1=308\cdot 1.7954545454545454=553$$

$$a_3=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{3-1}=308\cdot {1.7954545454545454}^2=308\cdot 3.2236570247933884=992.8863636363636$$

$$a_4=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{4-1}=308\cdot {1.7954545454545454}^3=308\cdot 5.787929658151765=1782.6823347107436$$

$$a_5=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{5-1}=308\cdot {1.7954545454545454}^4=308\cdot 10.39196461349976=3200.725100957926$$

$$a_6=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{6-1}=308\cdot {1.7954545454545454}^5=308\cdot 18.65830010151093=5746.756431265367$$

$$a_7=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{7-1}=308\cdot {1.7954545454545454}^6=308\cdot 33.50012972771281=10318.039956135546$$

$$a_8=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{8-1}=308\cdot {1.7954545454545454}^7=308\cdot 60.1479601929389=18525.57173942518$$

$$a_9=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{9-1}=308\cdot {1.7954545454545454}^8=308\cdot 107.99292852823122=33261.82198669521$$

$$a_{10}=a_1·r^{n-1}=308\cdot {1.7954545454545454}^{10-1}=308\cdot {1.7954545454545454}^9=308\cdot 193.8963944029606=59720.08947611186$$