等差数列{an}，a1=1，前n项和Sn，S2n/Sn=4

Question

（1）求数列｛an｝的通项共识和Sn
（2）记bn=an*2^（n-1）求｛bn｝的前N项和Tn

Answer 1

S2n=2n+n*(2n-1)d
Sn=n+n(n-1)d/2
4Sn=4n+2(n^2-n)d
S2n/Sn=4 S2n=4Sn

4n+2d(n^2-n)=2n+(2n^2-n)d
整理，得
dn=2n
d=2
S2n=2n+n*(2n-1)d
Sn=n+n(n-1)d/2
4Sn=4n+2(n^2-n)d
S2n/Sn=4 S2n=4Sn

4n+2d(n^2-n)=2n+(2n^2-n)d
整理，得
dn=2n
d=2

an=1+(n-1)*2=2n-1

Sn=n(n+1)-n=n^2

bn=an*2^(n-1)=n^2*2^n/2

Answer 2

。。都还给老师了

Answer 3

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A=2
故通式AN=(N-1)A+1=(N-1)*2+1=2N-1
Sn=(1+2n-1)n/2=n^2

bn=(2n-1)*2^(n-1)
Tn=1*2^0+3*2+5*2^2+...+(2n-1)*2^(n-1)
2Tn=1*2^1+3*2^2+5*2^3+...+(2n-1)*2^n
Tn-2Tn=1*2^0+2[2+2^2+...+2^(n-1)]-(2n-1)*2^n
-Tn=1+2*2(1-2^(n-1))/(1-2)-(2n-1)2^n=1-4(1-2^n/2-(2n-1)2^n=1-4+2*2^n-(2n-1)2^n=-3+(3-2n)2^n
故Tn=3-(3-2n)2^n

Answer 4

由当n=1时可得 a2=3 所以d=2
an=2n-1 sn=n^2

第二问用错位相减法
自己算很简单的
我不想算了