In the known sequence an, A1 = 2, the sum of the first n terms is SN. For any n ≥ 2, 3sn-4, an, 2-3 / 2S (n-1) is an arithmetic sequence (1) (2) if BN satisfies BN = 3Sn, find the first n terms and TN of BN

In the known sequence an, A1 = 2, the sum of the first n terms is SN. For any n ≥ 2, 3sn-4, an, 2-3 / 2S (n-1) is an arithmetic sequence (1) (2) if BN satisfies BN = 3Sn, find the first n terms and TN of BN

1、
When n ≥ 2, 3sn-4, an, 2-3 / 2 × s (n-1) form an arithmetic sequence, so 2An = 3sn-4 + 2-3 / 2 × s (n-1)
Because an = SN-S (n-1), 2 [SN-S (n-1)] = 3sn-3 / 2 × s (n-1) - 2
Therefore, Sn + 1 / 2 × s (n-1) = 2, that is, 2Sn + s (n-1) = 4
because
2Sn＋S(n-1)＝4
2S(n-1)+S(n-2)＝4
The subtraction of the two formulas is: 2An + a (n-1) = 0, so an = - 1 / 2 × a (n-1), n ≥ 2
So, when n ≥ 2, an = 1 / 2 × (- 1 / 2) ^ (n-2)
When n = 1, A1 = 2
2、
S1＝2,
When n ≥ 2,
Sn
＝a1＋a2＋.＋an
＝2＋1/2＋1/2×(-1/2)＋.＋1/2×(-1/2)^(n-2)
＝2＋1/2×[1－(－1/2)^(n-1)]/(1＋1/2)
＝7/3－1/3×(－1/2)^(n-1)
So, Sn = 7 / 3-1 / 3 × (- 1 / 2) ^ (n-1), (n = 1,2,...)
So, BN = 3Sn = 7 - (- 1 / 2) ^ (n-1), so,
Tn＝b1＋b2＋...＋bn＝7n－2/3×[1－(－1/2)^n]
Solve the equations x1-2x2 + 3x3-x4 = 1,3x1-x2 + 5x3-3x4 = 2,2x1 + x2 + 2x3-2x4 = 3
The title is wrong
（1） X1-2x2+3x3-x4=1,
（2） 3x1-x2+5x3-3x4=2,
（3） 2x1+x2+2x3-2x4=3
（1）+（3）=3x1-x2+5x3-3x4=4
It conflicts with (2) 3x1-x2 + 5x3-3x4 = 2, so the title is wrong
There are only four unknowns
There is no unique solution to this equation
That is to say, the final solution of this kind of equations needs to use a unique quantity to represent the other three unknowns
As shown in the figure, D is a point on the side BC of the equilateral triangle ABC. The vertical bisector EF of ad intersects AB, AC at points e, F and FD respectively. The extension of AB intersects at point M,
The extension of de intersects the extension of Ca at point n. to prove an = DM
The vertical bisector EF of ad intersects AB and AC at points E and f respectively
∴FA=FD,EA=ED
∴∠FAD=∠FDA,∠EAD=∠EDA
∴∠FAD+∠EAD=∠FDA+∠EDA
That is ∠ FAE = ∠ FDE
∴∠FAM=∠FDN
In △ AMF and △ dnf
FA=FD
∠FAM=∠FDN
∠ AFM = ∠ DFN (same angle)
∴△AMF≌△DNF
∴FM=FN
∴FM-FD=FN-FA
∴DM=AN
Let AE = x, then the square of AE = ed (2-x) + the square of 2 = x is x = 1.5
If it is diamond, then AE = AF, if Da ⊥ EF is known, then BD = 4-2 radical 2 BF = 4-2 radical 2
Ask: can't understand
What is the decreasing function interval of function f (x) = x ^ 3-3x ^ 2 + 1?
f(x）=x^3-3X^2+1
f'(x)=3x^2-6x
=3x(x-2)
When x > 2, f '(x) > 0
When 0
Finding the first solution of derivative function ^ + 3f-1
Find the decreasing interval of function, then 3x ^ 2-6x + 1 〈 0
Solution
Given the first n terms of an and Sn = 2 ^ n-1, then A1 ^ 2 + A2 ^ 2 +... + an ^ 2 is equal to
It is known that Sn = 2 ^ n-1
Then A1 = S1 = 1
When n ≥ 2, an = SN-S (n-1) = (2 ^ n-1) - (2 ^ (n-1) - 1) = 2 ^ (n-1)
∴an=2^(n-1).
It is an equal ratio sequence with the first term of 1 and the common ratio of 2,
Then the sequence with the square of an as the general term is an equal ratio sequence with 1 as the first term and 4 as the common ratio
The square of a1 + the square of A2 + the square of A3 +. + the square of an is:
（1-4^n)/(1-4)=(4^n-1)/3.
Solving the non-homogeneous equations: 3x1 + 2x3-2x4; 3x3 + 2x3-2x4,,
Elementary transformation of the augmented matrix of the equation
1 -2 3 -1 1
3 -1 5 -3 4
2 1 2 -2 3
Reduce matrix to diagonal matrix
1 2 3 -1 1
0 5 -4 0 1
0 5 -4 0 1
The rank of the matrix is 2,
So the solution of the equation is x = K1 {1,1 / 8,0,1 / 4} '+ K2 {0, - 1,1,1}' K1, K2 is an arbitrary constant
G is the center of gravity of △ ABC, EF passes through point G, and EF / / BC, if BC = 21, find the length of EF
EF：BC=2：3
EF=14
The type of discontinuity of y = SiNx / | x | is jump discontinuity? Why? High number
Let f (x) be defined in U (XO), and XO be the discontinuous point of F (x), then if both left continuous f (x -) and right continuous f (x +) exist, but f (x -) ≠ f (x +), then XO is called the jumping discontinuity of F (x), which belongs to the first discontinuity
For example: the first n terms of the continuous sequence {an} and Sn satisfy the following conditions: an, Sn, sn-1 / 2 (n is greater than or equal to 2) are equal proportion sequence, and A1 = 1,
Find the first n terms and Sn of sequence {an}
(1) From Sn ^ 2 = an (sn-1 / 2), an = sn-sn-1 (n ≥ 2), we can get Sn ^ 2 = (sn-sn-1) (sn-1 / 2), that is, 2sn-1sn = sn-1-sn. We know that sn-1sn ≠ 0, the two sides of the above formula are divided by sn-1sn to get 1 / sn-1 / sn-1 = 2 {1 / Sn} is an arithmetic sequence with the first term of 1 and the tolerance of 2, 1 / Sn = 1 + 2 (n-1) = 2N-1, Sn = 1 / (...)
An / Sn = Sn / (sn-1 / 2)
That is, (sn-sn-1) / Sn = Sn / (sn-1 / 2)
The results show that 2Sn * sn-1 + sn-sn-1 = 0
That is 1 / Sn = 2 + 1 / sn-1
So {1 / Sn} is the arithmetic sequence with the first term of 1 and d = 2
So 1 / Sn = 1 + (n-1) * 2 = 2N-1
So Sn = 1 / (2n-1) (when n > = 2)
After testing, n = 1 is also consistent with the meaning of the question
So Sn = 1 / (2n-1)
3x1 + 4x2-5x3 + 7x4 = 0 2x1-3x2 + 3x3-2x4 = 0 4x1 + 11x2-13x3 + 16x4 = 0 7x1-2x2 + X3 + 3x4 = 0 to solve the equations
It's urgent
Coefficient matrix A=
3 4 -5 7
2 -3 3 -2
4 11 -13 16
7 -2 1 3
r1-r2,r3-2r2
>
1 7 -8 9
2 -3 3 -2
0 17 -19 20
7 -2 1 3
r2-2r1,r4-7r1
>
1 7 -8 9
0 -17 19 -20
0 17 -19 20
0 -51 57 60
r3+r2,r4-3r2,r2*(-1/17)
1 7 -8 9
0 1 -19/17 20/17
0 0 0 0
0 0 0 0
r1-7r2
1 0 -3/17 13/17
0 1 -19/17 20/17
0 0 0 0
0 0 0 0
The general solution of the equations is: C1 (3,19,17,0) + C2 (13,20,0, - 17)