这道题怎么做？（请写出过程）

S1=1,S2=-3/2，所以
a1 = 1
a2 = -5/2

Sn - S<n-2> = 3(-1/2)^(n-1)
S<n+1> - S<n-1> = 3(-1/2)^n

两式相减
(S<n+1> - Sn) - (S<n-1> - S<n-2>) = 3(-1/2)^n - 3(-1/2)^(n-1)
a<n+1> - a<n-1> = 9*(-1/2)^n

对于 n 为偶数情况
a3 - a1 = 9/2^2 = 9/4
a5 - a3 = 9/2^4 = 9/4^2
a7 - a5 = 9/2^6 = 9/4^3
……
a<n+1> - a<n-1> = 9/4^(n/2)

以上各等式相加
a<n+1> - a1 = 9*[1/4 + 1/4^2 + …… + 1/4^(n/2)]
a<n+1> - 1 = 9* (1/4)*[1 - 1/4^(n/2)]/(3/4) = 3*[1 - 1/4^(n/2)]
a<n+1> = 4 - 3/4^(n/2) = 4 - 3/2^n
其中n为偶数

对于 n 为奇数数情况
a4 - a2 = -9/2^3
a6 - a4 = -9/2^5
a8 - a6 = -9/2^7
……
a<n+1> - a<n-1> = -9*/2^n

以上各等式相加
a<n+1> - a2 = -9*(1/2^3 + 1/2^5 + …… + 1/2^n)
a<n+1> + 5/2 = -9 * (1/2^3) * {1 - 1/4^[(n-1)/2)]/(1 - 1/4) = (-3/2)*[1 - 1/2^(n-1)]
a<n+1> = -4 + 3/2^n