解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=38$$ 和公比：$$r=1.7894736842105263$$ 插入几何级数的公式：

$$a_1=38$$

$$a_2=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{2-1}=38\cdot {1.7894736842105263}^1=38\cdot 1.7894736842105263=68$$

$$a_3=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{3-1}=38\cdot {1.7894736842105263}^2=38\cdot 3.2022160664819945=121.6842105263158$$

$$a_4=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{4-1}=38\cdot {1.7894736842105263}^3=38\cdot 5.730281382125675=217.75069252077566$$

$$a_5=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{5-1}=38\cdot {1.7894736842105263}^4=38\cdot 10.254187736435417=389.65913398454586$$

$$a_6=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{6-1}=38\cdot {1.7894736842105263}^5=38\cdot 18.349599107305483=697.2847660776083$$

$$a_7=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{7-1}=38\cdot {1.7894736842105263}^6=38\cdot 32.83612471833613=1247.772739296773$$

$$a_8=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{8-1}=38\cdot {1.7894736842105263}^7=38\cdot 58.759381074917286=2232.856480846857$$

$$a_9=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{9-1}=38\cdot {1.7894736842105263}^8=38\cdot 105.14836613406251=3995.6379130943756$$

$$a_{10}=a_1·r^{n-1}=38\cdot {1.7894736842105263}^{10-1}=38\cdot {1.7894736842105263}^9=38\cdot 188.1602341346382=7150.088897116251$$