The sum of the first n terms of the continuous sequence an is Sn = 10n-n ^ 2, the general term formula of the ball sequence {Ian I}, and the sum of the first n terms of {Ian I} and HN?

The sum of the first n terms of the continuous sequence an is Sn = 10n-n ^ 2, the general term formula of the ball sequence {Ian I}, and the sum of the first n terms of {Ian I} and HN?

According to the meaning of the title:
S(n-1)=10(n-1)-(n-1)^2
an=Sn-S(n-1)=11-2n
So: when n

For positive term sequence {an}, let HN = / (a1 + A2 / 2 + A3 / 3 + --- + an / N), if HN = 1 / (n + 1), then the general term formula of sequence an is
For positive term sequence {an}, let HN = (n + 1) / (a1 + A2 / 2 + A3 / 3 + --- + an / N), if HN = 1 / (n + 1), then the general term formula of sequence an is_____

According to the meaning of the question, (n + 1) / (a1 + A2 / 2 + A3 / 3 + --- + an / N) = 1 / (n + 1) so (n + 1) ^ 2 = a1 + A2 / 2 + A3 / 3 + --- + an / N let BN = an / N, B1 = A1 = 2 / H1 = 4, so (n + 1) ^ 2 = B1 + B2 + B3 +. + BN -------- let n = n + 1, so (n + 2) ^ 2 = B1 + B2 + B3 +... + BN + BN + 1 ---

Given the general term formula an = 31-3n of the sequence an, find the first n terms and HN of the sequence | an |
I know that when n ≥ 11, an ≤ 0, but why shouldn't HN = 2s10 Sn be equal to S10 + (- sn-s10) = - Sn

When n ≥ 11, an ≤ 0
So a11 + A12 +

Let {an} satisfy: a1 + 2A2 + 3a3 + +Nan = 2n (n ∈ n *) (1) find the general term formula of sequence {an}; (2) let BN = n2an, find the first n term and Sn of sequence {BN}

（1）∵a1+2a2+3a3+… +When Nan = 2n, 1, n ≥ 2, a1 + 2A2 + 3a3 + +(n-1) an-1 = 2N-1. ② ① - ② get Nan = 2N-1, an = 2n − 1n (n ≥ 2). Let n = 1 get A1 = 2, an = 2 (n = 1) 2n − 1n (n ≥ 2) (2) ∵ BN = 2 (n = 1) n · 2n − 1 (n ≥ 2). Then when n = 1, S1 = 2 ∵ when n ≥ 2, Sn = 2 + 2 × 2 + 3 × 22 + +N × 2N-1, then 2Sn = 4 + 2 × 22 + 3 × 23 + +By subtracting (n-1) · 2N-1 + n · 2n, Sn = n · 2n - (2 + 22 + 23 +) +In addition, it is proved that (2) Sn = 2n (2) ∈ Sn = 2N-1, Sn = 2n-2 ·

If a sequence {AI} satisfies A1 = 2, an + 1 = an + 2n (n is a natural number), then A100 is ()
A. 9900B. 9902C. 9904D. 10102

∵ A1 = 2, an + 1 = an + 2n (n is a natural number), ∵ A2 = 2 + 2 × 1, A3 = 2 + 2 × 1 + 2 × 2 = 2 + 2 × 3 An = 2 + n (n-1), ｜ A100 = 2 + 100 × (100-1) = 9902

It is known that A0 = 0, A1 = 1, an + 1 = 8an-an-1 (n = 1,2,.). In the sequence {an}, whether there are infinitely many terms divisible by 15. It is proved that

a(n+1)3^(n+1) = 3a(n)3^n + 1,
b(n)=a(n)3^n,
b(n+1) = 3b(n) + 1,
b(n+1) + x = 3[b(n) + x],1 = 3x-x=2x,x = 1/2.
b(n+1) + 1/2 = 3[b(n) + 1/2],
It is an equal ratio sequence whose first term is B (1) + 1 / 2 = a (1) * 3 + 1 / 2 = 7 / 2 and common ratio is 3
b(n)+1/2=(7/2)*3^(n-1)=(7/6)3^n,
a(n) = b(n)/3^n = 7/6 - (1/2)*(1/3)^n,n=1,2,...
LIM (n - > + infinity) a (n) = 7 / 6

Let {an} be an equal ratio sequence, A1 = 1000, q = 1 / 10, and BN = 1 / N (lga1 + lga2 +...) Find the maximum of the sum of the first n terms of {BN}

∵ {an} is an equal ratio sequence, A1 = 1000, q = 1 / 10 ∵ an = 1000 × (1 / 10) ^ (n-1) = 10 ^ (4-N) ∵ lgan = 4-N
∴bn＝1/n×[n×(3＋4－n)/2]＝(7－n)/2
When n ≥ 8, BN < 0
The largest sum of the first seven terms is 1 / 2 × 7 × (3 + 0) = 21 / 2

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. Find the maximum value of the sum of the first n terms of the sequence {BN} 2. Find the sum of the first n terms of the sequence {BN} Sn '& nbsp; & nbsp; (there should be a specific process & nbsp; & nbsp;

1.a1=1000,q=1/10,
=> an=a1×q^(n-1)=1000×(1/10)^(n-1)=10⁴/10^n,
=> bk=1/k(lga1+lga2+…… +lgak)=lg(a1×a2…… ×a3)/k
=lg[(10⁴/10¹)×(10⁴/10²)×(10⁴/103)×…… ×(10⁴/10^k)]/k
=lg[(10⁴)^k/10^(1+2+…… +k)]/k
=[4k-(1+k)×k/2]/k
=4-(1+k)/2
=> bn=4-(1+n)/2=(7-n)/2
=> Sn=(7-1)/2+(7-2)/2+…… +(7-n)/2
=7n/2-(1+2+…… +n)/2
=7n/2-n(n+1)/4
=(13n-n²)/4
=-[(n-13/2)²-(13/2)²]/4
Because n is a positive integer, when n is 6 or 7, the maximum value of Sn is 10.5
2.bn=(7-n)/2,
=>When N0; n > 7, BN ≤ 7, Sn '= Sn = (13n-n & # 178;) / 4,
n> 7, Sn '= S7 + | B8 | + | B9 | + +|bn|
=21/2+(8-7)/2+(9-7)/2+…… +(n-7)/2
=21/2+[(8+9+…… +n)-7×(n-7)]/2
=21/2+(n+8)×(n-7)/2-7×(n-7)]/2
=(n²-13n+84)/4
There is no problem with the method. Please check the calculation process
I hope I can help you_ ∩)o

In the sequence {an}, a1 + A6 = 33, a3a4 = 32, a (n + 1)

1.
a(n+1)

If {an} is an equal ratio sequence of positive terms, prove 1 / (lga1 * lga2) + 1 / (lga2 * lga3) +... + 1 / (LGA (n-1) * lgan) = (n-1) / (lga1 * lgan)

Let the common ratio of equal ratio sequence be Q
1/(lga1*lga2)+1/(lga2*lga3)+...+1/(lga(n-1)*lgan)
=1/[lga1(lga1+lgq)]+1/[(lga1+lgq)(lga1+2lgq)]+…… +1/[lga1+(n-2)q][lga1+(n-1)q]
={1/lga1-1/(lga1+lgq)+1/(lga1+lgq)-1/(lga1+2lgq)+…… -1/[lga1+(n-1)q]}/lgq
={1/lga1-1/[lga1+(n-1)q]}/lgq
=(n-1)/(lga1*lgan)