高二数列（答得好加分）

1.Sn = 2 - 2an;那么S(n+1) = 2 - 2a(n+1) ;两试一减得：
S(n+1) - Sn = (2-2a(n+1) - 2 + 2an) = 2an - 2a(n+1)
即：a(n+1) = 2an - 2a(n+1) 推出: a(n+1) = 2/3 *an;
由Sn = 2 - 2an得：a1 = 2 - 2a1;那么a1 = 2/3;
所以an 是等比数列，q = 2/3,a1 = 2/3;
2.an = (2/3)^n;
3.an*Sn = (2/3)^n * (2 - 2* (2/3)^n) = 2* (2/3)^n - 2*(4/9)^n
他的和为2*(1-(2/3)^(n+1))/(1-2/3) - 2*(1-(4/9)^(n+1))/(1-4/9)
= 12/5 - (2/3)^(n+1) + (4/9)^(n+1)