所以An=n
(2)Bn=n*p^n
可以利用错位相减来计算，具体为：
(1), Tn=1*p^1+2*p^2·······n*p^n
上式同乘p得：pTn= 1*p^2+2*p^3·······(n-1)*p^n+n*p^(n+1)
两式相减得(1-p)Tn=1*p^1+1*p^2·······1*p^n-n*p^(n+1)
显然，前n项为等差数列，求和，得Tn=p(1-p^n)/(1-p)^2-n*p^(n+1)
所以，答案是Tn=p(1-p^n)/(1-p)^2-n*p^(n+1)/1-p
(2)当P=1时，Tn=n（n+1）/2
当P不等于1时Tn=p(1-p^n)/(1-p)^2-n*p^(n+1)/1-p
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
简单分析一下，详情如图所示
...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...
② 由①-②得 (1-p)Tn=p+p^2+p^3+……+p^n-np^(n+1)=p(1-p^n)\/(1-p)-np^(n+1)所以Tn=[np^(n+1)]\/(p-1)+p(1-p^n)\/(1-p)²
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...
(1):因为数列{an}为等差数列,且a1=1,则由等差数列性质 可得:前n项和Sn=a1n-(n(n-1)\/2)*D 即Sn=n-(n(n-1)\/2)*D , S2n=2n-(2n(2n-1)\/2)*D 且 S2n\/Sn=(4n+2)\/(n+1),n=1,2,3```.(1),则将Sn,S2n代入(1)式,化简可得(2)式.因为(1)式对任意正整数都成立, ...
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...
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 ...
!!在等差数列{an}中,已知a1=1,前n项和Sn满足条件S
s(n) = n + n(n-1) = n^2.b(n) = a(n)2^[a(n)] = (2n-1)2^(2n-1) = (4n-2)2^(2n-2) = (4n-2)4^(n-1),t(n) = b(1) + b(2) + b(3) + ... + b(n-1) + b(n)= (4*1-2) + (4*2-2)4 + (4*3-2)4^2 + ... + [4(n-1)-2]...
等差数列{Sn}中,a1=1,前n项和Sn满足条件 S2n\/Sn=4, n=1,2,~~~.
a1+a2n =2a1+2an a2n = an + nd,其中d为等差公比；∴an +nd = a1 + 2an an = nd - a1 而：an = a1 + (n-1)d 对比：d = 2 所以：an = 2n-1 Sn = n^2 (2)bn = an*2^(n-1) = (2n-1) *2^(n-1)Tn = 2^0 + 3*2^1 + 5 * 2^2 + 7*2^3+......
等差数列{an},a1=1,前n项和Sn,S2n\/Sn=4,求其通项公式
没A1=1,A2=A+1,A3=2A+1...An=(N-1)A+1 所以SN=[(N-1)A+2]*N\/2 S2N=[(2N-1)A+2]*2N\/2 因为S2N=4SN 解得:-2A+4=-4A+8 所以2 故通式AN=(N-1)A+1=(N-1)*2+1=2N-1
...S2n\/Sn=4n+2\/n+1,An=na,是否存在正整数n和k,使Sn,S(n
由已知得S(2n)\/Sn=(4n+2)\/(n+1)=(4dn+2d)\/(dn+d)4a1-2d=2d，2a1-d=d 解得d=a1 an=na1，又已知an=na，因此a1=a Sn=na1=na 假设存在满足题意的n、k，则 S(n+1)²=Sn·S(n+k)[n(a+1)]²=(na)·[(n+k)a]整理，得(2a+1)n=ka²a=-1\/2时，...
等差数列 An S2n\/Sn=4n+2\/n+1,An=na,是否存在正整数n和k,使Sn,S(n
a1+a1=2-1 2a1=1 a1=1\/2 sn+an=2-(1\/2)^(n-1)s(n-1)+a(n-1)=2-(1\/2)^(n-2)两式相减得 2an-a(n-1)=(1\/2)^(n-2)-(1\/2)^(n-1)2an-a(n-1)=(1\/2)^(n-2)-1\/2*(1\/2)^(n-2)2an-a(n-1)=(1\/2)^(n-2)2an=a(n-1)+(1\/2)^(n-2)2a...
在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足:2Sn2=an(2Sn-1).(Ⅰ...
解答：（Ⅰ）证明：当n≥2时，其前n项和Sn满足：2Sn2=an（2Sn-1）．∴2S2n＝(Sn?Sn?1)(2Sn?1)，化为1Sn?1Sn?1=2，∴数列{1Sn}是等差数列，∴1Sn＝1+2(n?1)=2n-1，∴Sn=12n?1．（II）bn=Sn2n+1=1(2n?1)(2n+1)=12(12n?1?12n+1)，∴数列{bn}的前n项和为Tn=12...
