解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=1,200$$ 和公比：$$r=1.35$$ 插入几何级数的公式：

$$a_1=1200$$

$$a_2=a_1·r^{n-1}=1200\cdot {1.35}^{2-1}=1200\cdot {1.35}^1=1200\cdot 1.35=1620$$

$$a_3=a_1·r^{n-1}=1200\cdot {1.35}^{3-1}=1200\cdot {1.35}^2=1200\cdot 1.8225000000000002=2187.0000000000005$$

$$a_4=a_1·r^{n-1}=1200\cdot {1.35}^{4-1}=1200\cdot {1.35}^3=1200\cdot 2.4603750000000004=2952.4500000000007$$

$$a_5=a_1·r^{n-1}=1200\cdot {1.35}^{5-1}=1200\cdot {1.35}^4=1200\cdot 3.321506250000001=3985.8075000000013$$

$$a_6=a_1·r^{n-1}=1200\cdot {1.35}^{6-1}=1200\cdot {1.35}^5=1200\cdot 4.484033437500002=5380.840125000002$$

$$a_7=a_1·r^{n-1}=1200\cdot {1.35}^{7-1}=1200\cdot {1.35}^6=1200\cdot 6.053445140625002=7264.134168750003$$

$$a_8=a_1·r^{n-1}=1200\cdot {1.35}^{8-1}=1200\cdot {1.35}^7=1200\cdot 8.172150939843753=9806.581127812504$$

$$a_9=a_1·r^{n-1}=1200\cdot {1.35}^{9-1}=1200\cdot {1.35}^8=1200\cdot 11.032403768789068=13238.884522546881$$

$$a_{10}=a_1·r^{n-1}=1200\cdot {1.35}^{10-1}=1200\cdot {1.35}^9=1200\cdot 14.893745087865243=17872.49410543829$$