解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=10$$ 和公比：$$r=4.9$$ 插入几何级数的公式：

$$a_1=10$$

$$a_2=a_1·r^{n-1}=10\cdot {4.9}^{2-1}=10\cdot {4.9}^1=10\cdot 4.9=49$$

$$a_3=a_1·r^{n-1}=10\cdot {4.9}^{3-1}=10\cdot {4.9}^2=10\cdot 24.010000000000005=240.10000000000005$$

$$a_4=a_1·r^{n-1}=10\cdot {4.9}^{4-1}=10\cdot {4.9}^3=10\cdot 117.64900000000003=1176.4900000000002$$

$$a_5=a_1·r^{n-1}=10\cdot {4.9}^{5-1}=10\cdot {4.9}^4=10\cdot 576.4801000000002=5764.801000000002$$

$$a_6=a_1·r^{n-1}=10\cdot {4.9}^{6-1}=10\cdot {4.9}^5=10\cdot 2824.752490000001=28247.52490000001$$

$$a_7=a_1·r^{n-1}=10\cdot {4.9}^{7-1}=10\cdot {4.9}^6=10\cdot 13841.287201000006=138412.87201000005$$

$$a_8=a_1·r^{n-1}=10\cdot {4.9}^{8-1}=10\cdot {4.9}^7=10\cdot 67822.30728490003=678223.0728490003$$

$$a_9=a_1·r^{n-1}=10\cdot {4.9}^{9-1}=10\cdot {4.9}^8=10\cdot 332329.3056960102=3323293.0569601017$$

$$a_{10}=a_1·r^{n-1}=10\cdot {4.9}^{10-1}=10\cdot {4.9}^9=10\cdot 1628413.59791045=16284135.9791045$$