解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=51$$ 和公比：$$r=1.2352941176470589$$ 插入几何级数的公式：

$$a_1=51$$

$$a_2=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{2-1}=51\cdot {1.2352941176470589}^1=51\cdot 1.2352941176470589=63$$

$$a_3=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{3-1}=51\cdot {1.2352941176470589}^2=51\cdot 1.5259515570934257=77.82352941176471$$

$$a_4=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{4-1}=51\cdot {1.2352941176470589}^3=51\cdot 1.884998982291879=96.13494809688582$$

$$a_5=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{5-1}=51\cdot {1.2352941176470589}^4=51\cdot 2.3285281545958507=118.75493588438839$$

$$a_6=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{6-1}=51\cdot {1.2352941176470589}^5=51\cdot 2.8764171321478154=146.6972737395386$$

$$a_7=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{7-1}=51\cdot {1.2352941176470589}^6=51\cdot 3.5532211632414192=181.2142793253124$$

$$a_8=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{8-1}=51\cdot {1.2352941176470589}^7=51\cdot 4.389273201651165=223.85293328420943$$

$$a_9=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{9-1}=51\cdot {1.2352941176470589}^8=51\cdot 5.422043366745557=276.5242117040234$$

$$a_{10}=a_1·r^{n-1}=51\cdot {1.2352941176470589}^{10-1}=51\cdot {1.2352941176470589}^9=51\cdot 6.697818276568041=341.5887321049701$$