Find sequence, 6, 666 The sum of the first n terms

Find sequence, 6, 666 The sum of the first n terms

The general term formula is an = (2 / 3) (10 ^ n-1) = (2 / 3) 10 ^ n - (2 / 3)
sn=(20/27)(10^n-1)-(2/3)n

Sequence 6, 666 Sum up

The original formula = 2 / 3 * (9 + 99 + 999 + 99 9）
=2/3*[(10-1)+(10^2-1)+…… +(10^n-1)]
=2/3*[(10+10^2+…… +10^n)-n]
=2/3*[10*(1-10^n)/(1-10)-n]
=2/3*[10*(10^n-1)/9-n]

The sequence an satisfies A1 = A2 = A3 = 1, an + a (n + 1) + a (n + 2) + a (n + 3) = cos (n π / 2) to find s2014
SN is the sum of the first n terms of a sequence
There are several answers for reference, but please explain the process in detail
A -503 B 502 C -2 D 2

Grouping:
S2014=a1+a2+(a3+a4+a5+a6)+(a7+a8+a9+a10)+...+(a2011+a2012+a2013+a2014)
The rule in brackets: the first subscript of the nth bracket is 3 + 4 (n-1) = 4N-1, a total of (2011 + 1) / 4 = 503
In the k-th bracket:
a(4k-1)+a(4k-1+1)+a(4k-1+2)+a(4k-1+3)=cos[(4k-1)π/2]
=Cos (2k π - π / 2) = cos (- π / 2) = 0
S2014=a1+a2+0+0+...+0=a1+a2=1+1=2
Choose D

The common ratio Q of the equal ratio sequence {an} is known = - 12. (1) if A3 = 14, find the sum of the first n terms of the sequence {an}; (2) prove that for any k ∈ n +, AK, AK + 2, AK + 1 are equal ratio sequence

(1) From A3 = 14 = a1q2, and q = -12, we can get the first n items of the {{an} sequence, and Sn = 1 {[1 {[12) n] 1 + 12 = 2 {2 {(12) n 3 (2) proof: for any k ∈ = 14 = 14 = a1q2q2, and q = -12, we can get the first n items of the A1 = 1.1.the first n items of the sequence {{{{{an} {n}} and Sn = 1 [1 {[1 {[12 (12) n] 1} [1 {[1 {[1} [12) n] n] n] 1} [1 [1] 1 (1] 1 (1) n = 1.this (2) proof: proof: proof: proof: for any k ∈ K ∈ n} n, we can get 2q2q2q2q2q2q2q21 into arithmetic sequence

Let the sequence an satisfy: a (n + 1) = an ^ 2 - (Nan) + 1, and A1 = 2, find a general term of an

In fact, the general term formula of this sequence can not be solved because it depends on A1. However, since A1 has been given, there is a way to solve the mathematical induction. First of all, it can be calculated from the known conditions: A2 = A1 ^ 2-1a1 + 1 = 3a3 = A2 ^ 2-2a2 + 1 = 4a4 = A3 ^ 2-3a3 +

In the sequence {an}, A1 = 1, and for any positive integer n, there is an + 1 = an + N, then A100 = 1___ ．

∵a1=1，an+1=an+n，∴a2-a1=1，a3-a2=2，… ，a100-a99=99，∴a100=a1+（a2-a1）+（a3-a2）+… +（a100-a99）=1+1+2+… +99 = 4951. Answer: 4951

In the equal ratio sequence, A1 = 1, and for any natural number n, an-1 = an + N, then A100

a2=a1+1=1+1
a3=a2+2=1+1+2
a4=a3+3=1+1+2+3
...
an=an-1+n-1=1+1+2+3+..+(n-1)=1+(n-1)n/2
a100=1+100*99/2=4951

In the sequence {an}, A1 = 1, an + 1 = (1-1 / N + 1) an. If the inequality an + 1 + an + 2 +... + A2N > 1 / 12loga (A-1) + 2 / 3 holds for all natural numbers with n > 1, try to find the value range of A

An + 1 = (1-1 / N + 1) an then an + 1 = (n / N + 1) an then an + 1 = (n / N + 1) an = (n / N + 1) * (n-1 / N) an-1 =... = n / N + 1 * (n-1 / N) *.. 1 / 2 * A1 = 1 / N + 1, so an + 1 = 1 / N + 1 then an + 1 + an + 2 +... + A2N = 1 / N + 1 + 1 / N + 2 +... + 1 / N + n for 1 / N + 1 + 1 / N + 2 +... + 1 / N + N, let Sn = 1 / N + 1

Sequence an + 1 = an + (1 / 2) ^ n + 1, n belongs to natural number, and A1 = 1, let BN = 1 / 2an-3 / 4
Sequence an + 1 = an + (1 / 2) ^ n + 1, and A1 = 1, let BN = 1 / 2an-3 / 4
(1) Finding the general term formula of sequence {an}
(2) If CN = 2N-1, n belongs to natural number, find the first n terms and Sn of sequence {BN * CN}
(3) Under the condition of (2), if TN = (- 3N & # 178; - 2n + 3) / (2 (n + 1) &# 178;), n belongs to natural number, try to compare the size of Sn and TN, and explain the reason
The key point is the third question, the first two questions stay for me to proofread

I haven't learned mathematical induction yet. Do I have to use it?

The sum of the first n terms of the sequence {an} is Sn, A1 = 1, Sn = n ^ 2An (n is a natural number)

From Sn = n ^ 2An, s (n-1) = (n-1) ^ 2A (n-1) is obtained, and an = n ^ 2An - (n-1) ^ 2A (n-1) is reduced to an / a (n-1) = (n-1) / (n + 1) a (n-1) / a (n-2) = (n-2) / N Then an = A1 * 2 / [n (n + 1)] = 2 / [n (n + 1)]