Let BN = an + n prove that {BN} is an equal ratio sequence

Let BN = an + n prove that {BN} is an equal ratio sequence

S(n+1)=3Sn+n^2+2
Sn=3S(n-1)+(n-1)^2+2
therefore
a(n+1)=3an+2n-1
bn=an+n
b(n+1)=a(n+1)+(n+1)
therefore
b(n+1)/bn=[a(n+1)+(n+1)]/(an+n)
=[3an+2n-1+n+1]/(an+n)
=3
So it's an equal ratio sequence
Linear algebra, let A1, A2, A3 be the three solution vectors of 4-element non-homogeneous linear equations AX = B, and rank r (a) = 3, if A1 = [1,2,3,4] ^ t, 2a2-3a3 = [0,1, - 1,0] ^ t, then the general solution of the equations AX = B is "analysis"?
The basic solution system of R (a) = 3, ax = 0 has only one vector a (a1 + 2a2-3a3) = 0, so a1 + 2a2-3a3 = [1,3,2,4] ^ t is the non-zero solution of AX = 0, and the general solution of AX = B is k * [1,3,2,4] ^ t + [1,2,3,4] ^ t
In the triangle ABC, ad is perpendicular to BC be perpendicular to AC AD and be intersect at the point G ∠ ABC is equal to 45 degrees prove that BG times Ge is equal to Ag times GD
Right triangle age and bgd have an opposite vertex angle, and the two triangles are similar triangles
So: Ag: BG = Ge: Gd, that is, BG times Ge equals Ag times GD
Given that f (x) is an even function defined on R, then f (x) is a periodic function if and only if
If a is not equal to 0 and f (a + x) = f (A-X) holds for any x, I wonder why any a does not belong to R and f (a + x) = f (A-X) holds for any x, because if a is zero, we can also take other numbers and deduce that f (x) is a periodic function
If a is equal to 0, Let f (a + x) = f (A-X), that is, f (x) = f (- x), this is the definition of even function, and there is no periodicity! If a is not equal to 0, f (a + x) = f (A-X), even function f (A-X) = f (x-a), we get f (a + x) = f (x-a), and the period is 2A
Let the sum of the first n terms of the sequence an be Sn, if S1 = 1, S2 = 2, and Sn + 1-3sn + 2sn-1 = 0 (n > = 2), ask: is the sequence an equal ratio sequence
no
s3=3*s2-2*s1=4
a1=1
a2=1
a3=2
So it's not
In the arithmetic sequence an, if A3 + A4 + A5 + A6 + A7 = 450, then A2 + A8=______ .
From A3 + A4 + A5 + A6 + A7 = (A3 + A7) + (A4 + A6) + A5 = 5a5 = 450, A5 = 90, then A2 + A8 = 2a5 = 180
In the triangle ABC, CD / DA = AE / EB = 1 / 2, let BC vector = a, CA vector = B prove: de vector = 1 / 3 (B-A)
DE = AE - AD =(1/3) AB -(2/3) AC =(1/3)(- a - b )-(2/3)(- b )=(1/3)( b - a )
The following functions are discontinuous at the points indicated, indicating which class these discontinuous points belong to
y=x/tanx,x=kπ,x=kπ+π/2（k=0,±1,±2……）
The answer is that x = 0 and x = k π + π / 2 are removable discontinuities,
X = k π (K ≠ 0) is the second kind of discontinuity
Why? How?
When x = 0 and x = k π + π / 2, there is a limit for the function. It only needs to make the value of the function equal to 0, so it is a removable breakpoint
And x = k π (K ≠ 0) is the second kind of discontinuity
The denominator TaNx can't be 0? Isn't it true that the left and right limits of removable discontinuities exist and are equal
Let the sum of the first n terms of the sequence an be Sn, if S1 = 1, S2 = 2, and S (n + 1) - 3Sn + 2 (sn-1) = 0 (n > = 2), ask: is the sequence an equal ratio sequence
Can an = A1 * 2 ^ (n-1)
In the arithmetic sequence {an}, if A3 + A4 + A5 + A6 + A7 = 450, then the sum of the first nine terms of the sequence {an} is ()
A. 180B. 405C. 810D. 1620
The tolerance of ∵ sequence {an} is d ∵ from A3 + A4 + A5 + A6 + A7 = 450, we can get (a1 + 2D) + (a1 + 3D) + (a1 + 4D) + (a1 + 5d) + (a1 + 6D) = 450, we can get 5A1 + 20d = 450, that is, a1 + 4D = 90. Therefore, the sum of the first nine terms of sequence {an} is S9 = 9a1 + 9 × 82d = 9 (a1 + 4D) = 9 ×