等差数列an中,an=α,a2n=β,求a3n
在以d为公差的等差数列an中,设S1=a1+a2.+an,S2=an+1+an+2+a2n,S3=a2n+1+a2n+a
证明等差数列等差数列{an}中,证明[a1+a2+a3……+a2n-1]/(2n-1)=an注:分子上a2n-1中2n-
等差数列{an}中,a1=1,a2n=2an+1(n∈N+),Sn是数列{an}的前n项和,求an,Sn,(2)设数列{
在等差数列{an}中,a1+a3+a5+……+a2n-1=290,a2+a4+a6+……+a2n=261
设等差数列{An}的前n项和为Sn,S4=4S2,A2n=2An+1 ,(1)求数列{an}的通
已知等差数列{an}中,a1=2.an+1=an+3分之an 求an
等差数列An和等比数列Bn中a1=b1>0,a2n+1=b2n+1>0则
已知等差数列{an}中,a2=6,a5=15,若bn=a3n,则数列{bn}的前9项和等于______.
已知等差数列{an},若a2+a4+……a2n=a3a6,a1+a3+……=a2n-1=a3a5
已知数列{an}为等差数列,求证:{a3n+a3n-1}是等差数列.
已知等差数列{an}中,a2=6,a5=15,若bn=a2n,(1)求通项公式an.有一问a2=a1+d 6=a1+d
已知数列{an}是等比数列,{a2n-1}是等差数列,且a1+a2=18,求数列{an}的通项公式.