作业帮等差数列A1=1,前 n项和满足S2n/Sn=4n+2/n+1 设Bn=（An）p^(An),求前n项和
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/11 01:57:05
等差数列A1=1,前 n项和满足S2n/Sn=4n+2/n+1 设Bn=（An）p^(An),求前n项和
因为S2n=2n(a1+a2n)/2=n[2a1+(2n-1)d] ,Sn=n(a1+an)/2=n[2a1+(n-1)d]/2
又S2n/Sn=4n+2/n+1,所以[2+(2n-1)d]/[2+(n-1)d]=(2n+1)/(n+1)对任意正整数n都成立,解得d=1,于是An=n,Bn=np^n,
(1)当p=1时,Bn前n项和为Tn=n(n+1)/2
(2)当p≠1时 ,B1=p^1,B2=2p^2 ,B3=3p^3 ,……,Bn=np^n ,相加得：
Tn=p^1+2p^2 +3p^3 +…+np^n,乘以p得：pTn=p^2+2p^3 +3p^4 +…+np^(n+1)
错位相减得：（1-p)Tn=p^1+p^2 +p^3 +…+p^n-np^(n+1)
=p(1-p^n)/(1-p)-np^(n+1)
所以Tn=p(1-p^n)/(1-p)^2-np^(n+1)/(1-p)
又S2n/Sn=4n+2/n+1,所以[2+(2n-1)d]/[2+(n-1)d]=(2n+1)/(n+1)对任意正整数n都成立,解得d=1,于是An=n,Bn=np^n,
(1)当p=1时,Bn前n项和为Tn=n(n+1)/2
(2)当p≠1时 ,B1=p^1,B2=2p^2 ,B3=3p^3 ,……,Bn=np^n ,相加得：
Tn=p^1+2p^2 +3p^3 +…+np^n,乘以p得：pTn=p^2+2p^3 +3p^4 +…+np^(n+1)
错位相减得：（1-p)Tn=p^1+p^2 +p^3 +…+p^n-np^(n+1)
=p(1-p^n)/(1-p)-np^(n+1)
所以Tn=p(1-p^n)/(1-p)^2-np^(n+1)/(1-p)
等差数列A1=1,前 n项和满足S2n/Sn=4n+2/n+1 设Bn=（An）p^(An),求前n项和
等差数列an中,a1=1前n项和Sn,满足条件S2n/Sn=4n+2/n+1,求an通项 记Bn=anp^an(p>0)
等差数列an中,a1=1前n项和Sn,满足条件S2n/Sn=4n+2/n+1,求an通项
等差数列,a1=1,前n项和满足S2n/Sn＝(4n+2)/(n+1) n属于正整数 求an数列
在等差数列{an}中,a1=1,前n项和sn满足s2n/sn=4,n=1,
等差数列｛an｝中,a1=1,前n项和Sn满足条件S2n/Sn=4,n=1,2.,记bn=an*2^(n-1),求数列｛
在等差数列an中,a1=1,前N项和SN满足条件s2n/sn=4n+2/n+1,n=1,2,3.
等差数列{an},a1=1,前n项和Sn,S2n/Sn=4
在等差数列{an}中,a1=1,前n项的和sn满足条件S2n/S2=(4n+2)/(n+1),n=1,2.
等差数列{an}a1=1前n项和为Sn且S2n/Sn=4n+2/n+1 (1)求an通项公试
已知等差数列{an}满足 a1+a(2n-1)=2n设Sn是数列{1/an}的前n项和,记f(n)=S2n-Sn(n属于
在等差数列{AN}中,A1=1,前N项和SN满足条件S2N/SN=4N+2/N+1,N=1,2,…….求数列{AN}的通