解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=38$$ 和公比：$$r=1.2894736842105263$$ 插入几何级数的公式：

$$a_1=38$$

$$a_2=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{2-1}=38\cdot {1.2894736842105263}^1=38\cdot 1.2894736842105263=49$$

$$a_3=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{3-1}=38\cdot {1.2894736842105263}^2=38\cdot 1.6627423822714682=63.184210526315795$$

$$a_4=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{4-1}=38\cdot {1.2894736842105263}^3=38\cdot 2.1440625455605775=81.47437673130194$$

$$a_5=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{5-1}=38\cdot {1.2894736842105263}^4=38\cdot 2.764712229801797=105.05906473246829$$

$$a_6=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{6-1}=38\cdot {1.2894736842105263}^5=38\cdot 3.5650236647444227=135.47089926028806$$

$$a_7=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{7-1}=38\cdot {1.2894736842105263}^6=38\cdot 4.597004199275703=174.6861595724767$$

$$a_8=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{8-1}=38\cdot {1.2894736842105263}^7=38\cdot 5.927715941171301=225.25320576450943$$

$$a_9=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{9-1}=38\cdot {1.2894736842105263}^8=38\cdot 7.643633713615626=290.45808111739376$$

$$a_{10}=a_1·r^{n-1}=38\cdot {1.2894736842105263}^{10-1}=38\cdot {1.2894736842105263}^9=38\cdot 9.856264525451728=374.53805196716564$$