若数列an满足a1=1,an+a(n+1)=(1/4)^n ,设Sn=a1+4a2+4^2a3+……+4^（n-1）

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/11/09 05:13:15

若数列an满足a1=1,an+a(n+1)=(1/4)^n ,设Sn=a1+4*a2+4^2*a3+……+4^（n-1）*an,
类比课本中推导等比数列前几项和公式的方法,可求得5*Sn-4^n*an=___.

Sn=a1+4a2+4^2a3...+4^(n-1)an
4Sn=4a1+4^2a2+4^3a3...+4^nan
5Sn=[a1+4a2+4^2a3...+4^(n-1)an]+[4a1+4^2a2+4^3a3...+4^nan]
=a1+{4(a1+a2)+4^2(a2+a3)+...+4^(n-1)*[a(n-1)+an]}+4^nan
=(1+4^nan)+{4*(1/4)^1+(4^2)*(1/4)^2+...+[4^(n-1)]*[(1/4)^(n-1)]}
=1+4^nan+(n-1)*1=4^nan+n
于是5Sn-4^nan=n

若数列an满足a1=1,an+a(n+1)=(1/4)^n ,设Sn=a1+4*a2+4^2*a3+……+4^（n-1）设数列{an}的前n项和Sn=n^2-4n+1,则/a1/+/a2/+/a3/+……+/a10/= 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 已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1 1.数列{an}满足a1+a2+……+an=[4-(-2)^n]/3,n∈N*,则1/a1+1/a3+1/a5+……1/ 设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2) 设数列{an}满足a1+3 a2+3^2 a3+……+3^n-1 an=n/3,a属于N* 求数列{an}的通项数列{an}满足：1/a1+2/a2+3/a3+…+n/an=2n 设数列{an}满足a1+3a2+3^2a3+.3^n-1×an=n/3,a∈N+. 已知数列an满足a1=1,an=a1+2a2+3a3+4a4+.(n-1)a(n-1),求通项an 已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N 已知数列{an}满足a1=4,an+1=an+p.3^n+1(n属于N+,P为常数）,a1,a2+6,a3成等差数列.