Question

数学第六题求解要过程

Answer 1

如果题目是：
a1= 1 , Sn = 2a(n+1)
求Sn
则可以：
S(n+1) = 2a(n+2)
S(n+1)- Sn = 2a(n+2) - 2a(n+1) = a(n+1)
2a(n+2) = 3a(n+1)
a(n+2) : a(n+1) = 3:2 = q
S1 = a1 = 2a2=1 a2 = 1/2
an是从第二项a2开始的首项为1/2 , 公比是3/2的等比数列。
Sn = a1 + a2 * [q^(n-1) - 1]/(q-1)
=1 + 1/2 * [(3/2)^(n-1) -1]/(3/2 - 1)
= 1 + (3/2)^(n-1) -1
= (3/2)^(n-1)
故选：B