The N-1 power {BN} of an and Sn = 2-1 / 2 is the arithmetic sequence A1 = B1, A2 * (b2-b1) = A1 to find the general term of BN? Let CN = BN / an to find the general term of CN

The N-1 power {BN} of an and Sn = 2-1 / 2 is the arithmetic sequence A1 = B1, A2 * (b2-b1) = A1 to find the general term of BN? Let CN = BN / an to find the general term of CN

N-1 power of Sn = 2-1 / 2
an=sn-sn-1=-(1/2)^(n-2)
a1=-2
a2=-1
a2*（b2-b1）=a1
a2*（d）=a1
D=2
bn=1+(n-1)*2=2n-1
cn=bn/an=(2n-1)/[-(1/2)^(n-2)]
TN = sum of dislocation
Jiao Yiwei
For the linear equation system ax1 + x2 + X3 = 1 X1 + AX2 + X3 = a X1 + x2 + AX3 = a * a, when we ask what the value of a is, the equation system has unique solution or infinite solutions
For linear equations
aX1+X2+X3=1
X1+aX2+X3=a
X1+X2+aX3=a*a
For example, when we ask the value of a, do the equations have unique solutions or infinite solutions?
The augmented matrix is λ 1111 λ 1111 λ ^ 2. Firstly, the determinant of the coefficient matrix is calculated as λ 1111 λ = (λ + 2) (λ - 1) ^ 2. When λ ≠ 1 and λ ≠ - 2, the unique solution is known by Cramer's rule. When λ = 1, the augmented matrix is 1111 - > 1111 000
If the point m is the center of gravity of △ ABC, then the following vectors are collinear with the vector ab
A.AB+BC+AC
B.AM+MB+BC
C.AM+BM+CM
D.3AM+AC
Vector symbols
Choose C
A. AB + BC + AC = 2Ac, not collinear with ab
B. Am + MB + BC = AC, not collinear with ab
Let the midpoint of BC side be D, because m is the center of gravity, so am = 2Md, so 3am = 2ad
C: According to the parallelogram rule of vector addition, MB + MC = 2Md = am, so am + BM + cm = 0, the angle between zero vector and any vector is arbitrary, so AB and am + BM + cm are collinear
D: So the addition rule of 3AB = 2Ac + is not collinear with 3AB
D should be - 3AM, right?
Let AX = 3am and be collinear with am, by be parallel and equal to AC, then ABYC is a parallelogram, ay passes through the midpoint of BC, ay coincides with ax. So the vector AB + by = ax, that is, the vector AB + (- 3am + by) = 0, AB and (- 3am + AC) are collinear
Find the discontinuous point of function f (x) = (x-3) / (x ^ 3-x ^ 2-6x), and explain the type
How can I not do this kind of problem
One
It is known that the sequence {an} is an arithmetic sequence, A1 = 1, a1 + A2 + a3 = 12. Let BN = 3 ^ an, find the first n terms and Sn of the sequence {BN}
a1+a3=2a2
So 3a2 = 12
a2=4
Then d = a2-a1 = 3
an=3n-2
So BN = 3 ^ (3n-2)
Then B (n + 1) / BN = 3 ^ (3N + 1) / 3 ^ (3n-2) = 3 ^ 3 = 27
So BN is proportional, q = 27
b1=3^1=3
So Sn = 3 * (1-27 ^ n) / (1-27)
=3(27^n-1)/26
bn=9n-6 sn=4.5n^n-1.5n
If {an} is known to be an arithmetic sequence, then a1 + a3 = 2A2;
a1+a2+a3=3a2=12;
a2=4;a1=1;d=a2-a1=3;
an=1+3(n-1）=3n-2;
bn=3^an;bn=3^(3n-2)
bn=1/9*(27^n)
sn=1/9*(27^1)+1/9*(27^2)+。。。 +1/9*（27^n)
27sn=1/9*(27^2)+......+1/9*(27^n)+1/9*(27^n+1)
Just make a difference
Solution
26sn=3(27^n -1）
。
The linear equation system ax1-x2-x3 = 1 X1 + AX2 + X3 = 1 - X1 + x2 + AX3 = 0 has a unique solution
A necessary and sufficient condition for the existence of a unique solution of a system of non-homogeneous linear equations is as follows:
Rank of coefficient matrix = rank of augmented matrix = n (where n = 3)
It is not necessary to have 3 determinant equations because there are 3 solutions
Coefficient determinant=
a -1 -1
1 a 1
-1 1 a
= a^3-a
= a(a-1)(a+1).
So a ≠ 0 and a ≠ 1 and a ≠ - 1
As shown in Figure 2, △ ABC, ab = AC, ∠ BAC = 90 °, D is the midpoint of BC, de ⊥ DF, if be = 12, CF = 5, find the length of EF
Analysis:
If e is on AB, f is on AC and ad is connected, then
AD=(1/2)BC=DC,∠EAD=∠FCD=45°,∠EDA=90°-∠FDA=∠CDF,
∴△ADE≌△CDF,
∴AE=CF,
Similarly, AF = be,
AF = be = 12, AE = CF = 5
Also, EAF = 90 degrees
∴EF=√(AE²+AF²)=13
Point out the discontinuity of function f (x) = x - 3 / x ^ 2 - 9 and explain the reason
The function should be (x-3) / (x ^ 2-9), right? There are two discontinuities: x = 3, x = - 3,
When x tends to 3, the function is rewritten as 1 / (x + 3), and the limit is equal to 1 / 6, so x = 3 is the first kind of removable discontinuity point of the function. If f (3) = 1 / 6, the function is continuous at this point,
When x tends to - 3, the function tends to infinity, so x = - 3 is the second kind of infinite discontinuity of the function
In addition, this function is not called exponential function. The form of exponential function is a ^ X
In the sequence an, A1 = A2 = 1, and a (n + 2) = a (n + 1) + an, it is proved by mathematical induction that a 5N can be divided by 5
a5=5
Let n = k hold, that is, a 5 K can be divided by five (K ∈ n),
Then A5 (K + 1) = A5 + 4 + A5 + 3 = 2 * A5 + 3 + A5 + 2 =
=5*a5k+1 + 3*a5k
=5*i+5*j (i,j∈N)
That is, n = K + 1 holds
This is the famous "Fibonacci" sequence, Baidu search "Fibonacci" you will know the answer
The sequence of leby rabbit is proved
What conditions must a and B satisfy when homogeneous linear equations ax1 + x2 + X3 = 0, X1 + bx2 + X3 = 0, X1 + 2bx2 + X3 = 0 have nonzero solutions
There are nonzero solutions for homogeneous linear equations if and only if the rank of coefficient matrix is less than the number of unknowns
That is, the rank of the coefficient matrix is less than 3, that is, it is not full rank
The matrix is not full rank if and only if the value of the determinant of the matrix is equal to 0
We get B - A, B = 0
The solution is b = 0
Or B is not equal to zero, a = 1