解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=53$$ 和公比：$$r=1.528301886792453$$ 插入几何级数的公式：

$$a_1=53$$

$$a_2=a_1·r^{n-1}=53\cdot {1.528301886792453}^{2-1}=53\cdot {1.528301886792453}^1=53\cdot 1.528301886792453=81$$

$$a_3=a_1·r^{n-1}=53\cdot {1.528301886792453}^{3-1}=53\cdot {1.528301886792453}^2=53\cdot 2.335706657173372=123.79245283018871$$

$$a_4=a_1·r^{n-1}=53\cdot {1.528301886792453}^{4-1}=53\cdot {1.528301886792453}^3=53\cdot 3.569664891151757=189.19223923104312$$

$$a_5=a_1·r^{n-1}=53\cdot {1.528301886792453}^{5-1}=53\cdot {1.528301886792453}^4=53\cdot 5.455525588364006=289.14285618329234$$

$$a_6=a_1·r^{n-1}=53\cdot {1.528301886792453}^{6-1}=53\cdot {1.528301886792453}^5=53\cdot 8.337690050141218=441.89757265748455$$

$$a_7=a_1·r^{n-1}=53\cdot {1.528301886792453}^{7-1}=53\cdot {1.528301886792453}^6=53\cdot 12.742507435121484=675.3528940614386$$

$$a_8=a_1·r^{n-1}=53\cdot {1.528301886792453}^{8-1}=53\cdot {1.528301886792453}^7=53\cdot 19.474398155563023=1032.1431022448403$$

$$a_9=a_1·r^{n-1}=53\cdot {1.528301886792453}^{9-1}=53\cdot {1.528301886792453}^8=53\cdot 29.762759445294435=1577.426250600605$$

$$a_{10}=a_1·r^{n-1}=53\cdot {1.528301886792453}^{10-1}=53\cdot {1.528301886792453}^9=53\cdot 45.48648141639339=2410.7835150688497$$