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b

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/09/30 09:37:52

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+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

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b 设数列{an}的前n项和为Sn,并且满足2Sn=an²+n,an>0.(1)求a1,a2,a3.(2)猜想{a 设数列{an}的前n项和为Sn,满足2Sn=an+1-2^n+1+1,且a1,a2+5.a3成等差数列设数列{an}的前n项和为Sn,满足2Sn=an+1-2^n+1+1,且a1,a2+5.a3成等差数设数列an的前n项和为Sn,满足2Sn=an－2∧n+1 +1 ,且a1,a2+5,a3成等差设数列{an}满足Sn=n^2+1,Pn=1/a1.a2+1/a2.a3+.+1/an.an+1,求n=? 已知数列（an)的前N项和为SN,且满足sn=2an-n (n属于N+) 求（1）求a1 a2 a3 (2)求AN得通项设数列{an}的前n项和为Sn,满足2Sn=an+1-2^(n+1)+1,且a1,a2+5.a3成等差数列,求数列{an 数列an满足sn=3an-1/2 计算a1,a2,a3,a4 猜an通项求an前n项和sn 设数列{an}的前n项和为sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*) 已知数列an满足a1+2a2+3a3+...+nan=n(n+1)*(n+2),则数列an的前n项和Sn=? 已知数列{an}a1=2前n项和为Sn 且满足Sn Sn-1=3an 求数列{an}的通项公式an