已知数列an满足a1=0且Sn+1=2Sn+1/2n(n+1) 1,求 a2 a3,并证明an+1=2an+n 2,设bn=an+1-an,求证bn+1=2bn+

S(n+1)=2Sn+(1/2)n(n+1)1,求 a2 a3,并证明a(n+1)=2an+nS2=2S1+1=1a2=s2-a1=1S3=2S2+3=2+3=5a3=s3-s2=5-1=4所以a2=1,a3=4;S(n+1)=2Sn+(1/2)n(n+1)Sn=2S(n-1)+(1/2)n(n-1)两式相减：a(n+1)=2an+(1/2)n(n+1)-(1/2)n(n-1...