Extend the line segment AB to C so that BC = 2Ab, and then extend Ba to d so that ad = 3AB. Find the relationship between DC and AB, DC and BC, BD and AB, BD and BC!

All the segments are on a straight line, and DC = 6ab, DC = 3bC, BD = 4AB, BD = 2BC

The first n terms of the equal ratio sequence [3 ^ n] and Sn are equal to?

First item A1 = 3, common ratio = 3
Sn=a1(1-q^n)/(1-q)
=3(1-3^n)/(1-3)
=[3^(n+1)-3]/2

Let the sum of the first n terms of an equal ratio sequence be SN. If S6 is equal to S3 1:2, what is the ratio of S9 to S3

S6=a1(1-q^6)/(1-q)S3=a1(1-q^3)/(1-q)S6/S3=(1-q^6)/(1-q^3)=1/2(1+q^3)(1-q^3)/(1-q^3)=1/21-q^3=1/2q^3=1/2S9=a1(1-q^9)/(1-q)S9/S3=(1-q^9)/(1-q^3)=(1-q^3)(1+q^3+q^6)/(1-q^3)=1+q^3+q^6=1+1/2+1/4=7/4

The sum of the first n terms, the first 2n terms and the first 3N terms of the equal ratio sequence is Sn, s2n and s3n respectively. It is proved that Sn, s2n Sn and s3n-s2n are also equal ratio sequence

Sn=a1(1-q^n)/(1-q)
S2n=a1（1-q^2n）/(1-q)
S3n=a1(1-q^3n)/(1-q)
（S2n-Sn）/Sn=q^n
(S3n-S2n)/S2n-Sn=q^n
So these three are in an equal proportion sequence

If the sum of the first n terms, the first 2n terms and the first 3N terms of the equal ratio sequence is Sn s2n s3n respectively, prove Sn Λ 2 + s2n Λ 2 = Sn (s2n + s3n)

an = a1q^(n-1)Sn = a1(q^n-1)/(q-1)(Sn)^2 + (S(2n))^2= [a1(q^n-1)/(q-1)]^2 +[a1(q^(2n)-1)/(q-1)]^2= [a1/(q-1)]^2 .[ q^(4n) -q^(2n) - 2q^n + 2]Sn[S(2n) + S(3n) ]=[a1(q^n-1)/(q-1)] .[ a1(q^(2n)-1)/(q-1) ...

If Sn = 48, s2n = 60, then s3n=______ ．

∵ sequence {an} is an equal ratio sequence, and its first n terms and Sn, s2n Sn, s3n-s2n also form an equal ratio sequence. ∵ (60-48) 2 = 48 × (s3n-60), the solution is s3n = 63, so the answer is 63

What is the relationship between Sn, s2n, s3n of the sum of the first n terms of the equal ratio sequence?
Are the three numbers in equal proportion? If so, what is the common ratio of Q?

Let {an} and Sn, s2n, s3n have the following relations: Sn, s2n Sn, s3n-s2n form an equal ratio sequence, and the common ratio is Q ^ n. proof: first, a more general formula is proved. In the equal ratio sequence, an = a1q ^ (n-1) am = a1q ^ (m-1) is divided to obtain an / am = q ^ (n-m), an = AMQ ^ (n-m)

In the proportional sequence, Sn, s2n Sn, s3n-s2n are proportional
So in 1,2,4,8,16,32, S2 = 2, S4 = 8, S6 = 32
S2=2,S4-S2=6,S6-S4=24
S4-s2 / S2 is not equal to s6-s4 / s4-s2

Sn refers to the first n terms and the sequence S2 = 1 + 2 = 3s4 = 1 + 2 + 4 + 8 = 15, s4-s2 = 12s6 = 1 + 2 + 4 + 8 + 16 + 32 = 63, s6-s4 = 48, obviously 3,12,48 are proportional to each other. In fact, Sn = A1 + A2 +. + ans2n Sn = an + 1 + an + 2 +. + a2ns3n-s2n = A2N + 1 + A2N + 2 +. + a3n

If Sn = 25, s2n = 100, then s3n =?

325
In the equal ratio sequence, SK1: SK2: SK3 is equal to the K power of the original common ratio (k is a segment, that is, the ordinal number is from K to K, and then from K to 2K)
So SK1 = 25. SK2 = 75, so SK3 = 225
So s3n = 25 + 75 + 225 = 325
It should be like this, right!

It is known that each item of the equal ratio sequence {an} is a positive number, Sn = 80, s2n = 6560, and in the first n terms, the largest term is 54. Find the value of n

If q = 1, then there is a contradiction between Sn = Na1 = 80, s2n = 2Na1 = 160 and s2n = 6560, so Q ≠ 1. ∵ A1 (1 − QN) 1 − q = 80 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) & nbsp; A1 (1 − q2n) 1 − q = 6560 & nbsp; & nbsp; (2), then QN = 81 (3) ∵

Extend the line segment AB to C so that BC = 2Ab, and then extend Ba to d so that ad = 3AB. Find the relationship between DC and AB, DC and BC, BD and AB, BD and BC!