等差数列{an},a1=1,前n项和Sn,S2n/Sn=4

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

。。都还给老师了

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...

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