A 1=2, q= 1/2 So: sn = 2 * (1/2) (N- 1) (n is greater than or equal to1) Note: it is n-1.

(1) let G = n+ 1, then if n=g- 1 is substituted into sn, SG = s (n+1) = 2 * (12) (n+/kloc-

S (n+1) = (1/2) * 2 * (1/2) (n-1) = (1/2) * sn (n is greater than or equal to 0).

So s(n+ 1)=( 1/2)*sn.

② because a1>; 0, q>0, so sn>0, similarly, s(n+ 1)>0, so s (k+1)/(s (k)-c) > 2 is available. 2S(k)-2C so there is s (k+1)-2s (k) >; -C.

=》2*( 1/2)^n-2*2*( 1/2)^(n- 1)>； -C.

On the left, we can get 2 * (1/2) n (1-4) =-6 * (1/2) n =-6 * (1/2) n =-3 * (65438. =>H & gt-C (equivalent substitution)

Because H = -3 * (1/2) (n- 1), we can see that H is a geometric series with a common ratio greater than 0, and because the first term is -3 less than 0, H is increasing function. So when n takes 1, there is a minimum value of -6, that is, there is always H >；; =-6, so let H & gt-C always be true, then -C must be less than -6, and thus C >；; 6, so when c is a natural number greater than 6, (s (k+1)-c)/(s (k)-c) >; 2 forever.

After typing for so long, remember to add 100 to my promise.