Let positive number sequence {an} be equal ratio sequence, and A2 = 4, A4 = 16, find [LGA (n + 1) + LGA (n + 2) + +The value of the limit of LGA (2n)] / N ^ 2

Let positive number sequence {an} be equal ratio sequence, and A2 = 4, A4 = 16, find [LGA (n + 1) + LGA (n + 2) + +The value of the limit of LGA (2n)] / N ^ 2

According to the meaning of the question, we can get the common ratio q = 2, A1 = 2, and the general term an = 2 ^ n
The original formula = {LG [2 ^ (n + 1) * 2 ^ (n + 2) *... * 2 ^ (2n)]} / N ^ 2
={lg2^[(n+1)+(n+2)+...+(2n)]}/n^2
={[(n+1)+2n]*n/2}*lg2/n^2
When n tends to infinity, its limit is 3 / 2 * LG2

Let positive integer sequence an be an equal ratio sequence, and A2 = 4, A4 = 16, find LGA (n + 1) + LGA (n + 2) +. + LGA (2n)

Obviously an = 2 ^ n
lga(n+1)+lga(n+2)+.+lga(2n)
=lg(2^(n+1))+...lg(2^(2n))
=(n+1)lg2+...2n*lg2
=lg2*((n+1+2n)*n/2)
=lg2*((3n^2+n)/2)

In the arithmetic sequence with non tolerance, A5 = 7 and three numbers A1, A4 and A3 are in equal proportion sequence to find an

Is the tolerance not 0
A1, A4, A3 are equal ratio sequence
Let A1 = a
a4^2=a*a3
(a+3d)^2=a(a+2d)
a^2+6ad+9d^2=a^2+2ad
6ad+9d^2=2ad
4ad+9d^2=0
D is not equal to 0
a=-9d/4
a5=a+4d=(-9/4+4)d=(7/4)d=7
d=4
a=-9
So an = - 13 + 4N

In the arithmetic sequence an with non-zero tolerance, A5 = 7, and A1, A4 and A3 are in equal proportion sequence
1 find {an} 2 extract the first, second, second and #; 178 of the sequence {an};... 2 to form a new sequence {BN}, find the first n terms and Sn of the sequence {BN}

Let the tolerance of {an} be DA1, A4, A3, and the equal ratio sequence A4 & # 178; = A1 · A3 (a5-d) &# 178; = (a5-4d) (a5-2d) 4da5-7d & # 178; = 0a5 = 7. After sorting, we get D & # 178; - 4D = 0d (D-4) = 0d = 0 (rounding off with known contradiction) or D = 4A1 = a5-4d = 7-4 × 4 = - 9An = a1 + (n-1) d = - 9 + 4 (n-1) = 4n-13bn = a (...)