两道数列的题,求详细的过程和大致思路,看了解析以后还是不太明白……

an+1 -2 = -2an + 4 = -2 (an - 2)
即 （an - 2） 是等比数列,比例为 q = -2.
所以,an - 2 = (a1 - 2) * (-2)^(n-1)
化简后可以得到：an = 2 + 3*(-2)^(n-1)
a5 = a3 * q^(5-3) = 3 * q^2 = 27
所以,q = ±3
a1 = a3/q^2 = 1/3
所以,an = a1*q^(n-1) ,Sn = (an*q - a1)/(q - 1) = a1(q^n - 1)/(q - 1)
当 q = 3 时,an = 1/3 * 3^(n-1) = 3^(n-2),Sn = 1/3*(3^n - 1)/2 = 1/6 * (3^n - 1)
当 q = -3 时,an = 1/3 * (-3)^(n-1) = (-1)^(n-1)*3^n,Sn = 1/3 * [(-3)^n -1]/(-4) = -1/12*[(-3)^n - 1]