等差数列an中如果存在正整数k和L(k不等于L),使得前k项和Sk=k/L,前l项和SL=L/k,求Sk+L与4的关系,
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/05/14 05:33:51
等差数列an中如果存在正整数k和L(k不等于L),使得前k项和Sk=k/L,前l项和SL=L/k,求Sk+L与4的关系,
Sk=ka1+k(k-1)d/2=k/L
a1+(k-1)d/2=1/L (1)
SL=La1+L(L-1)d/2=L/k
a1+(L-1)d/2=1/k (2)
(1)-(2)
(k-L)d/2=1/L-1/k
(k-L)d/2 -(k-L)/kL=0
(k-L)[(d/2)-1/(kL)]=0
已知k≠L,k-L≠0,要等式成立,则d/2=1/(kL)
代入(1)
a1=1/L-(k-1)(d/2)=1/L-(k-1)/(kL)=1/(kL)
S(k+L)=(k+L)a1+(k+L)(k+L-1)(d/2)
=(k+L)/(kL)+(k+L)(k+L-1)/(kL)
=[(k+L)+(k+L)(k+L-1)]/(kL)
=(k+L)(1+k+L-1)/(kL)
=(k+L)²/(kL)
=(k²+2kL+L²)/(kL)
=(k/L)+(L/k)+2
k>0 L>0且k≠L,由均值不等式得k/L+L/k>2
S(k+L)>4
S(k+L)是大于4的.
a1+(k-1)d/2=1/L (1)
SL=La1+L(L-1)d/2=L/k
a1+(L-1)d/2=1/k (2)
(1)-(2)
(k-L)d/2=1/L-1/k
(k-L)d/2 -(k-L)/kL=0
(k-L)[(d/2)-1/(kL)]=0
已知k≠L,k-L≠0,要等式成立,则d/2=1/(kL)
代入(1)
a1=1/L-(k-1)(d/2)=1/L-(k-1)/(kL)=1/(kL)
S(k+L)=(k+L)a1+(k+L)(k+L-1)(d/2)
=(k+L)/(kL)+(k+L)(k+L-1)/(kL)
=[(k+L)+(k+L)(k+L-1)]/(kL)
=(k+L)(1+k+L-1)/(kL)
=(k+L)²/(kL)
=(k²+2kL+L²)/(kL)
=(k/L)+(L/k)+2
k>0 L>0且k≠L,由均值不等式得k/L+L/k>2
S(k+L)>4
S(k+L)是大于4的.
等差数列前n项和Sn Sm=k Sk=m 求Sm+k
已知数列{an}、{bn}都是等差数列,a1=-1,b1=-4,用Sk、Sk′分别表示数列{an}、{bn}的前k项和(
设Sn为等差数列{an}的前n项和,若a1=1,公差d=2,Sk+1-Sk=24,求K=多少?
设无穷等差数列{an}的前n项和为Sn.(Ⅰ)若首项a1=-4,公差d=2,求满足S(k^2)=(Sk)^2 的正整数k
设等差数列An的前n项和为Sn,若Sm=Sk=b则Sm+k=
设Sn为等差数列{an}的前n项和,若a1=1,a3=5,Sk+2-Sk=36,则K的值为
等比数列an中,前n项和Sn,则Sk,S2k-k,S3k-2k有何关系
设Sn为等差数列{an}的前n项和,若a1=1,公差d=2,Sk+2-Sk=24,则k=?
设Sn为等差数列{an}的前n项和,若a1=1,公差d=2,Sk+2-Sk=24,则k=( )
设无穷等差数列An的前n项和为Sn,若首项a1=3/2,公差d=1,求满足S(k的平方)=(Sk)的平方的正整数k
L+L+L=K+K K+K+K+K=O+O+O L+K+K+O=60 L+K+O=?
等差数列{an}前9项的和等于前4项的和.若a1≠0,Sk+3=0,则k=______.