問等比級數

Question

在等比級數〈an〉中，已知a1-a4=39，a2-a3=9，求公比

Answer 1

a1-a4=39→a1-a1r3=39→a1(1-r3)=39......(1)a2-a3=9→a1r-a1r2=9→a1(r-r2)=9........(2)(1)÷(2)→ (1-r3)/(r-r2)=39/99-9r3=39r-39r2→9r3-39r2+39r-9=0→(3r2-4r+1)(3r-9)=03(r-1)(3r-1)(r-3)=0→r=(3)或(1/3)或1(不合)答案：公比為3或1/3

Answer 2

設此等比級數為: a1 , a1*r , a1*r^2 , a1*r^3 --->"^"代表次方的意思
則: a1 - a1*r^3 = 39---(1) , a1*r - a1*r^2 = 9---(2)
由(1): a1 - a1*r^3 = 39 --->提出a1則得到下式
a1*(1 - r^3) = 39 --->再來利用(a^3-b^3)=(a-b)*(a^2+ab+b^2)這個公式
a1*(1 - r)*(1 + r + r^2) = 39---(3)
由(2): a1*r - a1*r^2 = 9 --->提出a1*r則得到下式
a1*r*(1 - r) = 9---(4)
(3) / (4):
a1*(1 - r)*(1 + r + r^2) = 39
-----------------------------------
a1*r*(1 - r) = 9

==>(1 + r + r^2) / r = 39 / 9 --->交叉相承後得到下式
9 + 9r + 9r^2 = 39r
9r^2 - 30r + 9 = 0
3r^2 - 10r + 3 = 0
(3r - 1)*(r - 3) = 0
r = 1/3 or 3

看不懂在問我= =