In the equal ratio sequence {an} (n ∈ n *), if A1 = 1, A4 = 18, then the sum of the first 10 terms of the sequence is () A. 2−128B. 2−129C. 2−1210D. 2−1211

In the equal ratio sequence {an} (n ∈ n *), if A1 = 1, A4 = 18, then the sum of the first 10 terms of the sequence is () A. 2−128B. 2−129C. 2−1210D. 2−1211

From A4 = a1q3 = Q3 = 18 {q = 12, so S10 = 1 − (12) 101 − 12 = 2 − 129

If A1 = 1, A4 = 1 / 8, then the sum of 10 items in the sequence is n +
The fourth power of A2-1 / 2 the second power of b2-1 / 2 the tenth power of C2-1 / 2 the eleventh power of D2-1 / 2

a4=a1q^3=q^3=1/8=(1/2)^3
q=1/2
S10=a1(1-q^10)/(1-q)
=1*(1-(1/2)^10)/(1-1/2)
=2(1-2^(-10))
=2-2^(-9)
No answer?
Is C 2 [1 - (1 / 2) ^ 10] Oh, if yes, choose C

In the equal ratio sequence {an}, A1 = 1, A4 = 1 / 8, find the general term formula of this sequence and the sum of the first 10 terms

a4=a1q^(n-1)=q^(4-1)=1/8
∴q=1/2
∴an=1/2^(n-1)
Sn=(1-1/2)/[1-(1/2)^n]

In the equal ratio sequence an, if A1 = 1, A4 = 1 / 8, what is the sum of the first 10 terms of the sequence

A1 = 1, A4 = 1 / 8, then d = - 7 / 24, Sn = DN ^ 2 / 2 + (a1 + D / 2) n, then S10 can be solved, as long as it is substituted in the front. I hope it can help the landlord
Please accept

In known sequence {an}, A1 = 1, Sn = 3A (n + 1) (1) find {an} general term an (2) find A2 + A4 + A6 +. + A2N note: n + 1 is subscript

1) Sn = 3A (n + 1) = > s (n-1) = 3an = > SN-S (n-1) = 3A (n + 1) - 3an = an, = > A (n + 1) = 4An / 3 = > {an} the first term A1 = 1, the equal ratio sequence of common ratio q = 4 / 3 = > an = A1 × Q & # 8319 / / Q = (4 / 3) ^ (n-1) 2) A2 + A4 + A6 + +a2n=(4/3)+(4/3)³+(4/3)^5+…… +(4/3)^(2n-...

If A1 = 1, 3A (n + 1) subscript = Sn (n belongs to n *), find the general formula of {an} and A2 + A4 + +A (2n) subscript

From 3an + 1 = Sn, we can deduce 3an + 2 = Sn + 1, then ② - ① can get 3an + 2-3an + 1 = an + 1, then we can easily deduce an + 2 / an + 1 = 4 / 3, then {an} is G.P +a2n=a2*(1-(q^2)^n)/1-q^2=12[(16/9)^...

{an} is an arithmetic sequence, A1 = 3. The N + 1 power of a subscript n * 2 - the n power of a subscript (n + 1) * 2 = the 2n power of 2

The N + 1 power of a subscript n * 2 - the n power of a subscript (n + 1) * 2 = the 2n power of 2
2 ^ n on both sides
an/2^(n-1)-a(n+1)/2^n=1
a1/2^0=3
So {an / 2 ^ (n-1)} is an arithmetic sequence with 3 as the first term and - 1 as the tolerance
an/2^(n-1）＝3－（n-1)=4-n
an=(4-n)*2^(n-1)
(n is n)
－－－－－－－－－－－－
Is {an} an arithmetic sequence?

Given the n-th power ((n ≥ 2, n ∈ n +) of A1 = 1, a ^ n = 2A ^ (n-1) (subscript) + 2 in the sequence {an}, find the general term formula of the sequence {an}

a(n)=2a(n-1)+2^n
Divide both sides by 2 ^ n
a(n)/2^n = a(n-1)/2^n-1 +1
Then a (n) / 2 ^ n is the arithmetic sequence with tolerance 1
a(n)/2^n=a(1)/2^1+(n-1)=n-1/2
a(n)=n*2^n-2^(n-1)
^Sign for power

It is known that the sequence {an} is an arithmetic sequence, and the sum of the first n terms is Sn, and A5 = 8, S5 = 20
It is known that the sequence {an} is an arithmetic sequence, the sum of the first n terms is Sn, and A5 = 8, S5 = 20 (1) find Sn; (2) if any n > t, n belongs to N, it has S1 + 2A1 + 1 / 6 + S2 + 2A2 + 1 / 6 +. + Sn + 2An + 1 / 6 > 12 / 25 to find the minimum value of T

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if A5 = 5A3, then s9s5=______ ．

∵ {an} is an arithmetic sequence, S9 = a1 + A2 + +a9=9a5，S5=a1+a2+… +A5 = 5A3, s9s5 = 9a55a3 = 9, so the answer is 9