If n > = 1, Sn = 2an-3, then an=

If n > = 1, Sn = 2an-3, then an=

3*2^(n-1)

The sum of the first n terms of the sequence {an} is Sn = 2-2an, n ∈ n *. Verification: the sequence {an} is an equal ratio sequence, and the general term an is obtained

It is proved that when n = 1, A1 = S1 = 2-2a1, ∧ A1 = 23, when n ≥ 2, an = sn-sn-1 = (2-2an) - (2-2an-1) = 2an-1-2an, ∧ Anan − 1 = 23. So {an} is an equal ratio sequence with A1 = 23 as the first term and q = 23 as the common ratio

Let the sum of the first n terms of the equal ratio sequence {an} be Sn, A1 = 2011, and an + 2An + 1 + an + 2 = 0 (n ∈ n *), then s2012?

Let the common ratio be Q
an+2an+1+an+2=0
∴ an+2an*q+an*q²=0
∴ an(1+2q+q²)=0
∵ an≠0
∴ 1+2q+q²=0
∴ (q+1)²=0
∴ q=-1
∴ S2012=a1+a2+.+a2012
=a1[1-(-1)^2012]/[1-(-1)]
=0

Let the sum of the first n terms of the equal ratio sequence {an} be Sn, A1 = 2011, and an + 2An + 1 + an + 2 = 0 (n ∈ n *), then s2012 = ()
A. 2011B. 2012C. 1D. 0

Let the common ratio of the equal ratio sequence {an} be q, ∵ an + 2An + 1 + an + 2 = 0, ∵ an + 2anq + anq2 = 0, that is, 1 + 2q + Q2 = 0, ∵ q = - 1, ∵ s2012 = A1 (1 − Q & nbsp; 2012) 1 − q = 2011 (1 − 1) 2 = 0, so D

The first n term of the sequence an is SN. It is known that 2an-2 ^ n = SN. It is proved that an-n · 2 ^ (n-1) is an equal ratio sequence

2an-2^n=sn
2a(n-1)-2^(n-1)=s(n-1)
If you want to reduce the two forms, there are some
2an-2a(n-1)-2^n+2^(n-1)=an
2an-2a(n-1)-2^(n-1)-an=0
an-2a(n-1)=2^(n-1)
an-n*2^(n-1)=2a(n-1)+2^(n-1)-n*2^(n-1)
an-n*2^(n-1)=2a(n-1)+(1-n)*2^(n-1)
an-n*2^(n-1)=2a(n-1)-(n-1)*2^(n-2)*2
an-n*2^(n-1)=2[a(n-1)-(n-1)*2^(n-2)]
therefore
An-n * 2 ^ (n-1) is an equal ratio sequence with a common ratio of 2
Hope to help you

If the sum of the first n terms of the equal ratio sequence is Sn, and am-1 + am + 1 - am ^ 2 = 0, s2m-1 = 38, then M = (M + 1, M-1, 2m-1 are all subscripts)

The sum of the first n terms of the arithmetic sequence {an} is SN. Given am-1 + am + 1 - (AM) ^ 2 = 0, s2m-1 = 38, find M
[interpretation]
Because am-1 + am + 1 = 2am
So 2am - (AM) ^ = 0
So am = 0 or am = 2
And s2m-1 = (a1 + a2m-1) * (2m-1) / 2 = 2am * (2m-1) / 2
That is, s2m-1 = am * (2m-1) = 38
It shows that am is not equal to 0, only am = 2
So 2m-1 = 19
So m = 10

In the known arithmetic sequence {an & nbsp;}, if an ≠ 0, and & nbsp; an-1-an2 + an + 1 = 0, the first (2n-1) term and s2n-1 = 38, then n is equal to ()
A. 10B. 19C. 20D. 38

The ∵ sequence {an} is an arithmetic sequence, ∵ 2An = an-1 + an + 1, an-1-an2 + an + 1 = 0, ∵ an (2-An) = 0, ∵ an ≠ 0, ∵ an = 2, s2n-1 = (2n − 1) (a1 + A2N − 1) & nbsp; 2 = (2n-1) an = 2 (2n-1) = 38, ∵ 2N-1 = 19, then n = 10

It is known that the first n terms of an and Sn satisfy Sn = k * 3 ^ 2-1 / 2 to find the value of K
(1) Finding the value of K
(2) Let BN = an (1 + log3an) find the first n terms of BN and TN
It is known that the first n terms of an and Sn satisfy Sn = k * 3 ^ n-1 / 2
It should be this

An=Sn-S(n-1)=(k*3^n-1/2)-(k*3^(n-1)-1/2)=k*(3^n-3^(n-1))
An is not equal to 0, so
q=An/A(n-1)=k*(3^n-3^(n-1))/k*(3^(n-1)-3^(n-2))=3
And A1 = S1 = k * (3 ^ 1-3 ^ (1-1)) = 2K
A2=S2-S1=k*(3^2-3^(2-1))-k*(3^1-3^(1-1))=6k-2k=4k
So q = A2 / A1 = 4K / 2K = 2
The title is still wrong

It is known that an is an equal ratio sequence, and the sum of the first n terms Sn = 5 ^ (n-1) + K, then k =?

Sn=5^(n-1)+kSn-1=5^(n-2)+kan=sn-sn-1=5^(n-1)+k-(5^(n-2)+k)=4/5*5^(n-1)a1=4/5 q=5Sn=(a1-an*q)/(1-q)=(4/5-4/5*5^(n-1)*5)/(1-5)=5^(n-1)-1/5Sn=5^(n-1)+kk=-1/5

Such as the title
Given the first n terms and Sn = a ^ n + k of the equal ratio sequence {an}, the general formula for finding an is
——It needs a process

a1=s1=a+k
Sn=(a+k)(1-q^n)/(1-q)=a^n+k
So (a + k) / (Q-1) = 1
(a+k)/(1-q)=k
q=a,k=-1,an=(a-1)a^(n-1)