S(n+1)=2Sn+(1/2)n(n+1)
1,求 a2 a3,并证明a(n+1)=2an+n
S2=2S1+1=1
a2=s2-a1=1
S3=2S2+3=2+3=5
a3=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)
=2an+n.
2,设bn=a(n+1)-an,求证b(n+1)=2bn+
a(n+1)=2an+n
a(n+1)+(n+1)=2an+2n+1
[a(n+1)+(n+1)]=2(an+n)+1
[a(n+1)+(n+1)+1]=2(an+n+1)
令cn=an+n+1
c(n+1)=2cn
cn=2^(n-1)c1=2^(n-1) (a1+1+1)=2^n
an+n+1=2^n
an=-n-1+2^n
bn=a(n+1)-an
=(2an+n)-an
=an+n
=(-n-1+2^n)+n
=-1+2^n
b(n+1)=-1+2^(n+1)
=-1+2*2^n
=-1+2*(bn+1)
=2bn +1
第2题简单方法：
bn=a(n+1)-an
=(2an+n)-an
=an+n
an=bn-n
a(n+1)=b(n+1)-(n+1)
b(n+1)-(n+1)=a(n+1)
=2an+n
=2(bn-n)+n
=2bn-n
b(n+1)-(n+1)=2bn-n
b(n+1)=2bn+1