解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=120$$ 和公比：$$r=2.25$$ 插入几何级数的公式：

$$a_1=120$$

$$a_2=a_1·r^{n-1}=120\cdot {2.25}^{2-1}=120\cdot {2.25}^1=120\cdot 2.25=270$$

$$a_3=a_1·r^{n-1}=120\cdot {2.25}^{3-1}=120\cdot {2.25}^2=120\cdot 5.0625=607.5$$

$$a_4=a_1·r^{n-1}=120\cdot {2.25}^{4-1}=120\cdot {2.25}^3=120\cdot 11.390625=1366.875$$

$$a_5=a_1·r^{n-1}=120\cdot {2.25}^{5-1}=120\cdot {2.25}^4=120\cdot 25.62890625=3075.46875$$

$$a_6=a_1·r^{n-1}=120\cdot {2.25}^{6-1}=120\cdot {2.25}^5=120\cdot 57.6650390625=6919.8046875$$

$$a_7=a_1·r^{n-1}=120\cdot {2.25}^{7-1}=120\cdot {2.25}^6=120\cdot 129.746337890625=15569.560546875$$

$$a_8=a_1·r^{n-1}=120\cdot {2.25}^{8-1}=120\cdot {2.25}^7=120\cdot 291.92926025390625=35031.51123046875$$

$$a_9=a_1·r^{n-1}=120\cdot {2.25}^{9-1}=120\cdot {2.25}^8=120\cdot 656.8408355712891=78820.90026855469$$

$$a_{10}=a_1·r^{n-1}=120\cdot {2.25}^{10-1}=120\cdot {2.25}^9=120\cdot 1477.8918800354004=177347.02560424805$$