In the sequence {an}, A1 = 1,3anan-1 + an-1 = 0 (n > = 2), it is proved that {1 / an} is an arithmetic sequence. Find the general term of the sequence

In the sequence {an}, A1 = 1,3anan-1 + an-1 = 0 (n > = 2), it is proved that {1 / an} is an arithmetic sequence. Find the general term of the sequence

There is a problem with the recursion, because both sides of the shift 3ana = - A are divided by a at the same time, so 3an = - 1 = = > an = - 1 / 3, so 1 / an = - 3 obviously {1 / an} is not an arithmetic sequence. The recursion should be 3ana + a-an = 0 (n ≥ 2). The shift a-an = - 3ana, then both sides are divided by ana at the same time, so 1 / an - 1 / a = - 3, so {1 / an} is

The known sequence an satisfies an = 1 + 2 +... + N, and 1 / A1 + 1 / A2 +... + 1 / an

an=1＋2＋3＋… If + n = [n (n + 1)] / 2, then: 1 / (an) = 2 / [n (n + 1)] = 2 [(1 / N) - 1 / (n + 1)], so: M = 1 / (A1) + 1 / (A2) + 1 / (A3) + ＋1/(an)=2[1/1－1/2]＋2[1/2－1/3]＋2[1/3－1/4]＋… In order to make mm {1 / N} - 1 / (n + 1)} = 2 {1-1 / (n + 1)}

The known sequence an satisfies an = 1 + 2 +... N, and (1 / A1) + (1 / A2) +... (1 / an)

an=1/[n(n+1)]=1/n-1/(n+1)
（1/a1）+(1/a2）+...(1/an)

In the known sequence {an}, for any positive integer n, there is always N2 = A1A2 If an Heng holds, then a1 + a3=______ ．

In ∵ sequence {an}, for any positive integer n, there is always N2 = A1A2 When n = 1, 1 = A1, when n = 2, 4 = A1A2, ∧ A2 = 4, when n = 3, 9 = a1a2a3, A3 = 94, ∧ a1 + a3 = 1 + 94 = 134, so the answer is: 134

In the arithmetic sequence {an}, A1 = 1, the first n terms and Sn satisfy the conditions s2nsn = 4, n = 1, 2 (1) Find the general term formula and Sn of sequence {an}; (2) note BN = an · 2N-1, find the first n term and TN of sequence {BN}

(1) Let the tolerance of the arithmetic sequence {an} be d. from s2nsn = 4, we can get: a1 + a2a1 = 4, so A2 = 3A1 = 3 and d = a2-a1 = 2, so an = a1 + (n-1) d = 2N-1, ｜ Sn = n (1 + 2n-1) 2 = N2; (2) from BN = an · 2N-1, we can get BN = (2n-1) · 2N-1 +（2n-1）•2n-1       ①2Tn=2+3•22+5•23+… +(2n-3) · 2N-1 + (2n-1) · 2n & nbsp; & nbsp; & nbsp; ② ① - ②: TN = 1 + 2.21 + 2.22 + +2•2n-1-（2n-1）•2n=2（1+2+22+… +2n-1）-（2n-1）•2n-1=2(1-2n)1-2-（2n-1）•2n-1∴-Tn=2n•（3-2n）-3．∴Tn=（2n-3）•2n+3．

In the known sequence {an}, A1 = 2, the sum of the first n terms is SN. When n = n * and N ≥ 2, there is always 3sn-4, an, 2 - (3 / 2) (sn-1), which is an arithmetic sequence
1. Find the general term formula of sequence {an}
2. If the sequence {BN} satisfies BN = Nan / 2, find the first n terms and TN of the sequence {BN}

1.An=2*(-1/2)^(n-1)
2. There are two understandings for ambiguity. The first one is that
N is a fixed value, TN = NSN / 2 = (n / 2) * 2 (1 + (- 1 / 2) ^ n) / 3 = n (1 + (- 1 / 2) ^ n) / 3
The second understanding is that n is not a fixed value, it may be 1,2,3,4 I can't figure it out

In the arithmetic sequence {an}, A1 = 1, the first n terms and Sn satisfy the conditions s2n / Sn = 4, n = 1,2. Note BN = an * 2 ^ (n-1), and find the first n terms of the sequence {BN}
Is to find the sum of the first n terms of the sequence {BN}!

S2/S1=(a1+a2)/a1=4,a1=1
So A2 = 3
Tolerance d = a2-a1 = 2
an=a1+(n-1)d=2n-1
So BN = (2n-1) * 2 ^ (n-1)
Let the sum of the first n terms of BN be TN
Then TN = 1 * 2 ^ 0 + 3 * 2 ^ 1 + 5 * 2 ^ 2 + +(2n-1)*2^(n-1)
2*Tn=1*2^1+3*2^2+5*2^3+…… +(2n-3)*2^(n-1)+(2n-1)*2^n
Subtracting the two formulas, we get - TN = 2 ^ 0 + 2 * 2 ^ 1 + 2 * 2 ^ 2 + +2*2^(n-1)-(2n-1)*2^n
=2^1+2^2+2^3+2^4+…… +2^n-(2n-1)*2^n-1
=2(1-2^n)/(1-2)-(2n-1)*2^n-1
=2*2^n-2-(2n-1)*2^n-1=(3-2n)*2^n-3
So the sum of the first n terms of the sequence {BN} is TN = 3 + (2n-3) * 2 ^ n

In the sequence {an}, A1 = 1 / 3, and for any n belonging to n *, n ≥ 2, an × an-1 = an-1-an holds
Let BN = 1 / an (n belongs to n *)
1: Finding the general term formula of sequence {BN}
2: Finding the first n terms and TN of sequence {an / N}

Analysis: 1. When n ≥ 2, an × a (n-1) = a (n-1) - an1 / an-1 / a (n-1) = 11 / an = 1 / a (n-1) + 1  sequence {1 / an} is an arithmetic sequence with 1 / A1 = 3 as the first term and D = 1 as the tolerance, 1 / an = 3 + (n-1) = n + 2An = 1 / (n + 2) BN = 1 / an = n + 22, let {an / N} be: CN, then: CN = 1 / N (n + 2) = 1 / 2 [1 / n-1 / (...)

For example, A1 496 and b928 fill in B1 4 C1 6 D1 9 and B2 2 C2 8 D2 9 with formulas respectively

A1 496 B14 =left(a1,1) c16 =mid(a1,2,1) d19 =right(a1,1)
B1 928 B22 =LEFT(B1,1) C28 =MID(B1,2,1) D29 =RIGHT(B1,1)

In sheet 1, if A1 is "20", B1 is "40", A2 is "15", B2 is "30", enter the formula "= a1 + B1" in C1, and change the formula
Copy from C1 to C2, and then copy the formula to D2, then the value of D2 is ()

C1 = a1 + B1, copy C1 to C2 (formula), then C2 = A2 + B2, copy C2 formula to D2, then D2 = B2 + C2 = B2 + (A2 + B2) = 75;
Because the formula you enter is relative to the cell, the result of evaluating the cell depends on the change of the corresponding letter column and number row;