解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=35$$ 和公比：$$r=1.2857142857142858$$ 插入几何级数的公式：

$$a_1=35$$

$$a_2=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{2-1}=35\cdot {1.2857142857142858}^1=35\cdot 1.2857142857142858=45$$

$$a_3=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{3-1}=35\cdot {1.2857142857142858}^2=35\cdot 1.6530612244897962=57.85714285714287$$

$$a_4=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{4-1}=35\cdot {1.2857142857142858}^3=35\cdot 2.125364431486881=74.38775510204084$$

$$a_5=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{5-1}=35\cdot {1.2857142857142858}^4=35\cdot 2.732611411911704=95.64139941690965$$

$$a_6=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{6-1}=35\cdot {1.2857142857142858}^5=35\cdot 3.513357529600763=122.9675135360267$$

$$a_7=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{7-1}=35\cdot {1.2857142857142858}^6=35\cdot 4.517173966629553=158.10108883203435$$

$$a_8=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{8-1}=35\cdot {1.2857142857142858}^7=35\cdot 5.8077950999522825=203.2728284983299$$

$$a_9=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{9-1}=35\cdot {1.2857142857142858}^8=35\cdot 7.467165128510078=261.3507794978527$$

$$a_{10}=a_1·r^{n-1}=35\cdot {1.2857142857142858}^{10-1}=35\cdot {1.2857142857142858}^9=35\cdot 9.600640879512959=336.02243078295356$$