1, because the sequence {√Sn} is a arithmetic progression with a tolerance of D. So the general term of the sequence {√Sn} can be written as √ sn = n * d+t. So S 1=(d+t)? ，S2=(2d+t)？ ，S3=(3d+t)？ . So a 1=S 1=(d+t)? ，a2=S2-S 1=(2d+t)？ -(d+t)？ =d(3d+2t)，a3=S3-S2=(3d+t)？ -(2d+t)？ =d(5d+2t). Because 2a2=a 1+a3, 2d(3d+2t)=(d+t)? +d(5d+2t), t=0, so the general term of sequence {√Sn} is √Sn=n*d, so Sn=n? *d？ , so a 1=S 1=d? When n≥2, an = sn-s (n-1) = (2n-1) d? , so the general term of the sequence {an} is an=(2n- 1)d? .

2. Make Sm+Sn & gt；; C*Sk holds for any positive integer m, n and k, so let m? +n？ & gtc*k？ Do c for any positive integer m, n, k.