1.已知等差数列(an)的前n项和为Sn,bn=1/Sn,且a3+a5=21 2已知等比数列{an}满足an>0,n=1,2等,且a5*a(2n-5)=2的2n次方,则当n≥1时log2a1+log2a3+..+log2a(2n-1)=多少 3.an=2n-1 .bn的前n项和为Tn=1-1/2bn,若cn=(3的n次方*bn)/an
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 15:51:42
![1.已知等差数列(an)的前n项和为Sn,bn=1/Sn,且a3+a5=21 2已知等比数列{an}满足an>0,n=1,2等,且a5*a(2n-5)=2的2n次方,则当n≥1时log2a1+log2a3+..+log2a(2n-1)=多少 3.an=2n-1 .bn的前n项和为Tn=1-1/2bn,若cn=(3的n次方*bn)/an](/uploads/image/z/14399337-57-7.jpg?t=1.%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%28an%29%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Cbn%3D1%2FSn%2C%E4%B8%94a3%2Ba5%3D21+2%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3an%EF%BC%9E0%2Cn%3D1%2C2%E7%AD%89%2C%E4%B8%94a5%2Aa%282n-5%29%3D2%E7%9A%842n%E6%AC%A1%E6%96%B9%2C%E5%88%99%E5%BD%93n%E2%89%A51%E6%97%B6log2a1%2Blog2a3%2B..%2Blog2a%282n-1%29%3D%E5%A4%9A%E5%B0%91+3.an%3D2n-1+.bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%3D1-1%2F2bn%2C%E8%8B%A5cn%3D%283%E7%9A%84n%E6%AC%A1%E6%96%B9%2Abn%29%2Fan)
1.已知等差数列(an)的前n项和为Sn,bn=1/Sn,且a3+a5=21 2已知等比数列{an}满足an>0,n=1,2等,且a5*a(2n-5)=2的2n次方,则当n≥1时log2a1+log2a3+..+log2a(2n-1)=多少 3.an=2n-1 .bn的前n项和为Tn=1-1/2bn,若cn=(3的n次方*bn)/an
1.已知等差数列(an)的前n项和为Sn,bn=1/Sn,且a3+a5=21 2已知等比数列{an}满足an>0,n=1,2等,且a5*a(2n-5)=2的2n次方,则当n≥1时log2a1+log2a3+..+log2a(2n-1)=多少 3.an=2n-1 .bn的前n项和为Tn=1-1/2bn,若cn=(3的n次方*bn)/an*a(n+1),Sn为cn的前n项和,证明:Sn<1
1.已知等差数列(an)的前n项和为Sn,bn=1/Sn,且a3+a5=21 2已知等比数列{an}满足an>0,n=1,2等,且a5*a(2n-5)=2的2n次方,则当n≥1时log2a1+log2a3+..+log2a(2n-1)=多少 3.an=2n-1 .bn的前n项和为Tn=1-1/2bn,若cn=(3的n次方*bn)/an
1的一个是要问什么,2的一个根据等比数列的性质可得a5*a(2n-5)=2^2n=a1*a(2n-1),依次类推,原式可变为log2(a1×a3..×a(2n-3)×a(2n-1))=log2(2^2n×2^2n..×2^2n)一共乘以n/2个2^2n,你就可以知道答案了
3的一题应该把bn和cn做出来就好证明了
(1)已知数列an的前n项和为sn满足sn=an²+bn,求证an是等差数列(2)已知等差数列an的前n项和为sn,求证数列sn/n也成等差数列
设Sn为等差数列an的前n项和.求证Sn/n为等差数列
已知等差数列an中,前n项和sn=n^2-15n,则使sn为最小值的n
等差数列{an}的前n项和为sn,a10
设Sn为等差数列{An}的前n项和,求证:{Sn/n}是等差数列
等差数列an的前n项和为Sn,已知a5=11 a8=5求an和Sn
已知sn=32n-n^2求等差数列|an|的前n项和sn
sn为等差数列{an}的前n项的和,已知s15>0,s16
设等差数列{an}的前n项和为Sn,已知a3=11,S15*S16
已知等差数列{an} 的前n项和为Sn,若S12>0,S13
已知Sn为等差数列an的前n项和,a6=100,则S11=
已知等差数列An的前n项和为Sn,且S13>0,S14
设等差数列{an}的前n项和为Sn,已知S12>0,S13
已知Sn为等差数列{an}的前n项和,S11=55,则a6=
已知等差数列{an}的前n项和为Sn,S15>0,S16
已知等差数列{an}的前n项和为Sn,则S7-S3除以S10=
已知等差数列an的前n项和为Sn,且a1^2+a8^2
已知正项数列an的前n项和为Sn,a1=1,(an-2)²=8Sn-1.证明an是等差数列.