等差数列{Sn}中,a1=1,前n项和Sn满足条件 S2n/Sn=4, n=1,2,~~~~~.

Question

（1）求数列｛an}的通项公式an和前n项和Sn：

（2）记bn=an*2^(n-1),求数列｛bn}前n项和Tn：

Answer 1

解：
(1)S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)
Sn = n(a1+an)/2
S(2n)/Sn = 2(a1+a2n)/(a1+an) = 4
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+..............+(2n-1) *2^(n-1)
2Tn =2 + 3**2^2+5 * 2^3+............................+(2n-3)*2^(n-1)+(2n-1) *2^n
两式相减：
-Tn = 2^0 + 2*2 + 2*2^2 +...................+2^2^(n-1)-(2n-1) *2^n
Tn = -1+(2n-1) *2^n - 4(2^n -1)追问

为什么S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)？

Answer 2

(1)等差数列{Sn}中，S2n=2Sn+n^2*d，
所以S2n/Sn=(2Sn+n^2*d)/Sn=2+n^2*d/Sn=4
得到：Sn=n^2*d/2
因为S1=a1，所以d/2=1,得d=2
所以an=a1+(n-1)d=1+(n-1)*2=2n-1
Sn=n^2
(2)bn=an*2^(n-1)=(2n-1)*2^(n-1)=n*2^n-2^(n-1)
记{n*2^n} 前n项和为Pn ,{2^(n-1)}前n项和为Qn
Pn=1*2^1+2*2^2+3*2^3+......+n*2^n
2Pn= 1*2^2+2*2^3+......+(n-1)*2^n+n*2^(n+1)
相减得：-Pn=1*2^1+1*2^2+1*2^3+......+1*2^n-n*2^(n+1)=2^(n+1)-2-n*2^(n+1)=(1-n)*2^(n+1)-2
所以Pn=(n-1)*2^(n+1)+2
又因为Qn= 2^(n-1)-1
所以：Tn=Pn -Qn =(n-1)*2^(n+1)+2 - 2^(n-1)+1=(4n-5)*2^(n-1)+3

Answer 3

(1) S1=a1=1,S2=a1+(a1+d)=2+d
(2+d)/1=4
d=2
an=1+(n-1)*2=2n-1
(2) bn=an*2^(n-1)=(2n-1)*2^(n-1)

Tn=b1+b2+b3+--------+bn
=1*2^0+3*2^1+5*2^2+-------+(2n-1)*2^(n-1) (1)
2Tn= 1*2^1+3*2^2+-------+(2n-3)*2^(n-1)+(2n-1)*2^n (2)
(2)式-(1)式得
Tn=-2(2^1+2^2+--------+2^(n-1)+(2n-1)*2^n-1
化简得
Tn=（2n-3)*2^n+3