If the sum of the first n terms of the equal ratio sequence {an} is Sn = a-12n. (1) find the value of the real number a; (2) find the sum of the first n terms of the sequence {Nan}

If the sum of the first n terms of the equal ratio sequence {an} is Sn = a-12n. (1) find the value of the real number a; (2) find the sum of the first n terms of the sequence {Nan}

(1) When n = 1, A1 = S1 = A-12 (2 points) when n ≥ 2, an = sn-sn-1 = (a-12n) - (a-12n − 1 & nbsp;) = 12n (5 points) then A1 = 12 = A-12, the solution is a = 1 (7 points) (2) Nan = n2n, then RN = 12 + 222 + 323 + +n2n，①… (10 points) 2rn = 1 + 22 + 322 + +n2n−1，②… (11 points) ② - ①: RN = 2-N + 22n (15 points)

(an) is a sequence with common ratio Q, LQL > 1, BN = an + 1 (n = 1,2...), if the sequence (BN) has four consecutive terms in the set (- 53, - 23,19,37,82), then 6q=

{BN} has four consecutive terms in {- 53, - 23,19,37,82}
Bn=An+1 An=Bn-1
Then {an} has four consecutive terms in {- 54, - 24,18,36,81}
If {an} is an equal ratio sequence, and there is a negative number term in the equal ratio sequence, then Q1, in this case, should be omitted)
∴q=-1.5
∴6q=-9

Let BN = log2an (n ∈ n *), B1 + B3 + B5 = 6, b1b3b5 = 0. (I) find the general term formula of the sequence {an}; (II) let the sum of the first n terms of {BN} be Sn, if S11 + S22 + +The value of n when SNN is maximum

(I) B1 + B3 + B5 = log2 (a1a3a5) = log2 (a13aa5) = log2 (a13aa5) = log2 (a13aq6) = 6 {a13q6 = 6 {a13q6 = 26 {a1q2 = 4, \57a1
A1
A1
1, {B1 = log2a1 ≠ 0, and b1b3b5 = 0, if B3 = 0, then log2a3 = log2 (a1q2) = log2 (a1q2) = 0, that is, a1q2 = 4 = 4 = 4, so B5 = log2 (a1q44) = 0 {a1q4 = 1 {a1q44 = 1 {a1q44 = 1 {4 = 1 {4 = 1 {4 = 1 Q44 = 1 {q24 = 14, q = 14, q = 14, q = 14, q = 12, q = 12, = 5-N, {BN} is the first term of 4 with a tolerance of - 1 So {SNN} is an arithmetic sequence with the first term of 4 and the tolerance of - 12. When n ≤ 8, SNN > 0; when n = 9, SNN = 0; when n > 9, SNN < 0. So when n = 8 or n = 9, S11 + S22 + +SNN is the largest

Let {an} be an equal ratio sequence with common ratio Q, | Q | 1, let BN = an + 1 (n = 1, 2,...) If the sequence {BN} has four consecutive terms in the set {- 53, - 23, 19, 37, 82}, then 6q = ()
A. -9B. -3C. 9D. 3

Because BN = an + 1 (n = 1, 2,...) And the sequence {BN} has four consecutive terms in the set {- 53, - 23,19,37,82}. Therefore, an ∈ {- 54, - 24,18,36,81} because {an} is an equal ratio sequence with common ratio Q and | Q | 1, the terms in the sequence {an} are: - 24,36, - 54816q = 6 × (− 32) = − 9, so a is selected

The sequence {an} is an equal ratio sequence with the first term of 1000 and the common ratio of 1 / 10. The sequence {BN} satisfies BK = 1 / K (lga1 + lga2 +...) ＋lgak﹚k∈N*
(1) Finding the maximum value of the sum of the first n terms of the sequence {BN} (2) finding the sum of the first n terms of the sequence {BN} s'

an=a1q^(n-1)=1000×(1/10)^(n-1)=1/10^(n-4)
bk=(1/k)(lga1+lga2+...+lgak)
=(1/k)lg(a1×a2×...×ak)
=(1/k)lg[(1/10)^(1+2+...+k-4k)]
=-(1/k)lg[10^(k(k+1)/2-4k)]
=-(1/k)[k(k+1)/2 -4k]
=(7-k)/2
Sn=b1+b2+...+bn=(7/2)n -(1+2+...+n)/2
=(7/2)n-n(n+1)/4
=(-n²+13n)/4
=-(n-13/2)²/4 +169/16
When n = 6 and N = 7, the minimum value of (N-13 / 2) & # - 178 / 4 is - 1 / 16, and the maximum value of Sn is (SN) max = 21 / 2
two
Let BN ≥ 0
(7-n)/2≥0
N ≤ 7, that is, the first seven items of the sequence are all nonnegative, starting from the eighth item, and then each item is negative

Determine the convergence and divergence of the series ntan (π Λ n + 1)
For example, we determine the convergence and divergence of the series ntan (π 2 Λ n + 1), where π is divided by 2 to the power of N + 1

A:
limn->∞ u(n+1)/u(n)
=limn->∞ [(n+1)tan(π/2^(n+2))]/[ntan(π/2^(n+1))]
And when T - > 0, t ~ t
=limn->∞ [(n+1)(π/2^(n+2))]/[n(π/2^(n+1))]
=limn->∞ (n+1)/(n*2)
=1/2

The convergence and divergence of discriminant series ∑ (n + 1) / 2 ^ n
Judging the convergence and divergence of series ∑ (n + 1) / 2 ^ n, the range of summation is 1-n
Sum range 1 to n

The convergence of the series can be judged by the ratio criterion. For the sum, the power series is made
　　　f(x) = ∑{n>=0}(n+1)x^n,|x|=0}(n+1)∫[0,x](t^n)dt
　　= ∑{n>=0}x^(n+1)
　　= 1/(1-x) - 1,|x|

Find the convergence range of power series ∑ (n ^ 2 + 1) * x ^ n / (n! * 2 ^ n) and its sum function

﹙﹣∞,﹢∞﹚
[e^﹙x/2﹚]﹙1＋x/2＋x²/4﹚

Sum function ∑ (1 / 2n + 1) x ^ 2n + 1 (N 0 → positive infinity) of power series in convergence region

Let f (x) = ∑ (1 / 2n + 1) x ^ 2n + 1
The derivation f '(x) = ∑ x ^ 2n = 1 / (1-x ^ 2), the convergence domain is | X|

Seeking convergence interval sum function of power series X ^ (n + 1) / n
Σ = n is the front

ρ = LIM (n - > ∞) | [1 / (n + 1)] / (1 / N) | = LIM (n - > ∞) | (1 + n) | = 1 the convergence radius is r = 1 / ρ = 1
When x = 1, the ∑ [x ^ (n + 1)] / N = ∑ 1 / N series diverges
When x = - 1, the ∑ [x ^ (n + 1)] / N = ∑ [(- 1) ^ (n + 1) / N] series converges
So the convergence interval of power series ∑ x ^ (n + 1) / N is [- 1,1)
Let s (x) = ∑ x ^ (n + 1) / N = x ∑ (x ^ n) / N = - XLN (1-x) (- 1)