If the sum of the first n terms of the equal ratio sequence {an} is 2N-1, then the sum of the first n terms of the sequence {an ^ 2} is?

If the sum of the first n terms of the equal ratio sequence {an} is 2N-1, then the sum of the first n terms of the sequence {an ^ 2} is?

Is Sn = 2 ^ n-1?
S(n-1)=2^(n-1)-1
So an = SN-S (n-1) = 2 ^ (n-1)
So an ^ 2 = 4 ^ (n-1)
a1^2=1
So sum = 1 * (1-4 ^ n) / (1-4) = (4 ^ n-1) / 3

The relationship between the sum of the first n terms and the sum of the 2n terms of the equal ratio sequence

a(n)=aq^(n-1)
When q = 1,
s(n)=na,s(2n)=2na=2s(n).
When q is not equal to 1,
s(n)=a[q^n-1]/(q-1),s(2n)=a[q^(2n)-1]/(q-1)=a[q^n-1]/(q-1)*[q^n+1]=s(n)[q^n+1]

The sum of the first n terms is 2, and the sum of the last 2n terms is 12. What is the sum of the first 3N terms?

If the sum of items 1 to n is 2, and the sum of items n + 1 to 2n is 12-2 = 10, then the sum of items 2n + 1 to 3N is 10 * (10 / 2) = 50, and the sum of the first 3N is 2 + 10 + 50 = 62

In the arithmetic sequence {an}, A4 = 10 and A3, A6, A10 are equal ratio sequence, find the sum of the first 20 terms of the sequence {an} S20

Let the tolerance of sequence {an} be D, then A3 = a4-d = 10-d, A6 = A4 + 2D = 10 + 2D, A10 = A4 + 6D = 10 + 6D. From A3, A6 and A10, we can get a3a10 = A62, that is, (10-d) (10 + 6D) = (10 + 2D) 2, then we can get 10d2-10d = 0, and get d = 0 or D = 1. When d = 0, S20 = 20a4 = 200. When d = 1, A1 = a4-3d = 10-3 × 1 = 7, then S20 = 20a1 + 20 × 192D = 20 × 7 + 190 = 330

In the arithmetic sequence {an} with tolerance DD and the arithmetic sequence {BN} with common ratio Q, it is known that A1 = B1 = 1, A2 = B2. A8 = B3
There is also a question that a workshop produces ten pieces of a certain product every day in three days, and one or two defective products are produced on the first day and the second day respectively Find the probability of passing the inspection on the first day, and find the probability of passing all the previous two days

In the arithmetic sequence {an} with tolerance DD and the arithmetic sequence {BN} with common ratio Q, it is known that A1 = B1 = 1, A2 = B2. A8 = B3
a2=1+d
b2=q
q=1+d
a8=1+7d
b3=q^2=1+2d+d^2
d^2=5d
D = 0 or D = 5
Q = 1 or q = 6
1.an=1,bn=1
2.an=5n-4,bn=6^(n-1)
There is also a question that a workshop produces ten pieces of a certain product every day in three days, and one or two defective products are produced on the first day and the second day respectively Find the probability of passing the inspection on the first day
The probability of passing the inspection on the first day P1 = 9 / 10
p2=4/5
The probability of all passing in the first two days = P1 * P2 = 36 / 50

It is known that the sequence {an} is an arithmetic sequence with non-zero tolerance D, and the sequence {a (BN)} is an arithmetic sequence with common ratio Q, B1 = 1, B2 = 10, B3 = 46. Find the common ratio Q and BN

A1, A10, a46 are equal ratio sequence, ■ (a1 + 9D) ^ 2 = A1 (a1 + 45d)
Therefore, an = (n + 2) d
a1=3d,a10=12d,∴q=4.
abn=(bn+2)d=3dq^(n-1)
∴bn=3×4^(n-1)-2.

It is known that the sequence {an} is an arithmetic sequence with non-zero tolerance. A1 = 1, if A1, A2, A5 are equal ratio sequence, then an
a5=1+4d
a2=1+d
1+4d=(1+d)^2
d^2-2d=0
d≠0
d=2
an=1+2(n-1)=2n-1
Want to know why step 3 A5 = A2 ^ 2 is

Solution
∵ A1, A2, A5 are equal ratio sequence
∴a2²=a1a5
From A1 = 1
∴a2²=a5
∴(1+d)²=1+4d
∴1+2d+d²=1+4d
That is D & # 178; - 2D = 0
‖ d = 0 or D = 2
∵d≠0
∴d=2
∴an=1+(n-1)×2=2n-1

Let the sum of the first n terms of the sequence {an} with all positive numbers be s
Let the sum of the first n terms of the sequence {an} with all positive numbers be Sn, and the known sequence {Sn} is an arithmetic sequence with the first term of 1 and the tolerance of 1. Find the general formula of the sequence {an}

If {Sn} is an arithmetic sequence with the first term of 1 and the tolerance of 1, then
√Sn=1+1×(n-1)=n
So Sn = n ^ 2, an = sn-1-sn = n ^ 2 - (n-1) ^ 2 = 2N-1
General term formula 2N-1 of sequence {an}

The tolerance D ≠ 0 of the arithmetic sequence {an} is known, and A1, A3 and A9 are equal ratio sequence,
Then the value of (a1 + a3 + A9) / (A2 + A4 + A10) is?

You can get it from the question
A1 * A9 is equal to A3
Write the numerator and denominator as expressions of A3 and D
The relation between A3 and D can be obtained from the above formula
You can bring in the ratio

If the tolerance of arithmetic sequence an is not equal to 0 and A1, A3 and A9 are equal ratio sequence, divide (a1 + a3 + A9) by (A2 + A4 + A10)

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