Let an = n ^ 2 + λ n, A1

Let an = n ^ 2 + λ n, A1

Because A1 = 1 + λ, A2 = 4 + 2 λ
Since a1-3, the axis of symmetry is n = - λ / 2-3)
So an in n > - λ / 2（

In the sequence {an}, if A1 = 1 and A1 × A2... × an = n & sup2; for all n ∈ n +, then A3 + A5 =?

a1×a2...×an=n² ---- 1
Then A1 × A2... × an-1 = (n-1) & # 178; --- 2
Type 1 / 2
an=n²/(n-1)²
Then A3 = 9 / 4
a5=25/16
Then A3 + A5 = 61 / 16

In the sequence {an}, A1 = 1, a (n + 3) = an + 3, a (n + 2) ≥ an + 2 (n ∈ n *) (1) find A7, A5, A3, A6;

a(3)>=a(1)+2
a(5)>=a(3)+2
a(7)>=a(5)+2
a(7)>=a(1)+6
a(7)=a(4)+3=a(1)+3+3=a(1)+6
So wait for everything
a(3)=a(1)+2
a(5)=a(3)+2
a(7)=a(5)+2
a(3)=3
a(5)=5
a(7)=7
a(3)>=a(1)+2
a(5)>=a(3)+2
a(7)>=a(5)+2
a(9)>=a(7)+2
a(11)>=a(9)+2
a(13)>=a(11)+2
a(15)>=a(13)+2
a(17)>=a(15)+2
a(19)>=a(17)+2
a(19)>=a(1)+18
a(19)=a(16)+3=a(13)+6=a(10)+9=a(7)+12=19
So all of the above are equal
a(15)=a(12)+3=a(9)+6=a(7)+2=9
a(6)=a(9)-3=6

The arithmetic sequence an and the positive term sequence BN, and A1 = B1 = 1, A3 + A5 + A7 = 9, a7 is the equal proportion median of b3b7, the general formula for finding an and BN,

Should positive term sequence BN be positive term equal ratio sequence?
Because the sequence {an} is an arithmetic sequence, and A3 + A5 + A7 = 9, A5 = 3, and A1 = 1, then an = (n + 1) / 2,
So A7 = 4, because A7 is the middle term of b3b7, that is, B5 = A7 = 4, so the common ratio is √ 2, then BN = 2 ^ [(n-1) / 2]