等差数列{an},a1=1,前n项和Sn,S2n/Sn=4(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
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![等差数列{an},a1=1,前n项和Sn,S2n/Sn=4(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](/uploads/image/z/5619400-16-0.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%2Ca1%3D1%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%2CS2n%2FSn%3D4%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%B1%E8%AF%86%E5%92%8CSn%EF%BC%882%EF%BC%89%E8%AE%B0bn%3Dan%2A2%5E%EF%BC%88n-1%EF%BC%89%E6%B1%82%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E5%89%8DN%E9%A1%B9%E5%92%8CTn)
等差数列{an},a1=1,前n项和Sn,S2n/Sn=4(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
等差数列{an},a1=1,前n项和Sn,S2n/Sn=4
(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
等差数列{an},a1=1,前n项和Sn,S2n/Sn=4(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
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...
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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
收起
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