In the sequence {an} (subscript, the same after), A1 = 1, A2 = 2, the sequence {Anan + 1} is an equal ratio sequence with common ratio Q (Q > 0) Problem: find the first 2n terms and s2n of sequence {an}

In the sequence {an} (subscript, the same after), A1 = 1, A2 = 2, the sequence {Anan + 1} is an equal ratio sequence with common ratio Q (Q > 0) Problem: find the first 2n terms and s2n of sequence {an}

anan+1=q(an-1an)
That is, an + 1 = qan-1
That is to say, every two terms are equal
So look at the odd
a(2n+1)=a1*q^n
a(2n)=a2*q^(n-1)
S2n=a1+a2+…… a2n=(a1+a3+…… a2n-1)+(a2+…… A2N) respectively use the sum formula of equal ratio sequence
=（1-q^n)/(1-q)+2(1-q^n)/(1-q)
=3(1-q^n)/(1-q)

Given that the sequence {an} satisfies the relation LG (1 + A1 + A2 +. + an) = n, find the general term formula of the sequence

lg（1+a1+a2+.+an）=n
1+Sn=10^n
Sn=10^n-1
When n = 1, A1 = S1 = 9
When n ≥ 2, an = SN-S (n-1)
=10^n-10^(n-1)
=9*10^（n-1）
When n = 1, the above formula also holds
∴an=9*10^(n-1)

Given that the sequence {an} satisfies the relation LG (1 + A1 + A2 +. + an) = n, find the general term formula of the sequence {an}

The N-1 power of A1 = 10 and an = 9 * 10

In {an}, log2 ^ a (n + 1) = 1 + log2 ^ an, and a1 + A2 + +If A100 = 100, then A101 + A102 + +a200=?

By sorting log2 ^ a (n + 1) = 1 + log2 ^ an, we get that it is an equal ratio sequence, q = 10, a1 + A2 + +A100 = 100, so
a101+a102+…… +a200=q^100*(a1+a2+…… +a100)=10^102

It is known that the sequence an is composed of positive numbers A1 = 3, and satisfies lgan = lgan-1 + log C, where n is a positive integer and C > 0. Find the general term formula of the sequence an

Log C in the title should be LG C, otherwise the problem cannot be solved
lg an=lg an-1+lg c=lgc*an-1
an=c*an-1
an/an-1=c
C is a positive number and a constant at the same time, so we can judge that this sequence is equal ratio sequence Q = C
an=a1*q^(n-1)=3*c^(n-1)

Let an = 2n - 7 (n ∈ n +), then | A1 | + | A2 | + | A3 | + +|a15|=?

From an = 2n - 7 (n ∈ n +) we can get A1 = 2-7 = - 5A2 = 4-7 = - 3a3 = 6-7 = - 1A4 = 8-7 = 1, thus we can get that an is the arithmetic sequence of tolerance d = 2, and when n > = 4, an > 0, so | A1 | + | A2 | + | A3 | + +|a15|=|a1|+|a2|+|a3|+(a4+a5+…… +a15)=5+3+1+1*12+[12*(12-1)*2]/2=153...

In the sequence {an}, A1 = 3, an + an-1 + 2N-1 = 0 (n belongs to N and N > = 2) (1) find the value of A2 and A3

When n = 2, bring in the original formula: A2 + A1 + 3 = 0, A1 = 3, A2 = - 6N = 3, A3 + A2 + 5 = 0, A3 = 1, the original formula = (an + n) + (an-1 + n-1) = 0, so that (an + n) = BN, so BN + bn-1 = 0, that is BN = - bn-1 BN / bn-1 = - 1, so (an + n) is an equal ratio sequence of - 1, B1 = a1 + 1 = 4, BN = 4 * (- 1) ^ n + 1An + n =

In the sequence {an}, A1 = 1 / 3, and (a1 + A2 + a3 +...) +an)/n=(2n-1)an,(n∈N*)
(1) Write out the general term formula of the first five (2) inductive conjecture {an} of this sequence and prove it

(a1+a2+a3+… +An / N = (2n-1) an, then Sn = n (2n-1) an, or an = Sn / [n (2n-1] and S2 = a1 + A2 = 2 * 3 * A2 = 1 / 3 + a25a2 = 1 / 3 = > A2 = 1 / 15. Similarly, A3 = 1 / 35, A4 = 1 / 63a5 = 1 / 1023 and A1 = 1 / 3 = 1 / (4-1) = 1 / (2 ^ 2-1) A2 = 1 / 15 = 1 / (16-1) = 1 / (2 ^ 4-1) A3 = 1 / 35 = 1 / (36-1)

The sequence {an} satisfies A1 / 1 + A2 / 3 + A3 / 5 + +An / (2n-1) = 3 ^ (n + 1), then the general term formula of the sequence {an} is?

Let Sn = A1 / 1 + A2 / 3 + A3 / 5 + +An / (2n-1) = 3 ^ (n + 1), then: an = SN-S (n-1) where n > 1, n ∈ n + ｜ an / (2n-1) = 3 ^ (n + 1) - 3 ^ (n) = 2.3 ^ (n) where n > 1, n ∈ n + an = 2 (2n-1) · 3 ^ (n) so: 1 N = 1An = 2 (2n-1) · 3 ^ (n) n > 1

If A2 + A6 + a16 in an arithmetic sequence {an} is a definite constant, then ()
A. S17B. S15C. S8D. S7

∵ A2 + A6 + a16 = a1 + D + A1 + 5D + A1 + 15d = 3 (a1 + 7D) = 3A8 is a definite constant, while 2a8 = a1 + A15, ∵ S15 = 15 (a1 + A15) 2 = 15a8 = 5 × 3A8 is a definite constant