The known sequence {an} and {BN} satisfy: A1 = λ, an + 1 = 2 / 3an + n-4, BN = (- 1) ^ n (an-3n + 21), where λ is a real number and N is a positive integer 1. Prove that the sequence {an} is not equal ratio sequence for any real number λ 2. Prove that when λ≠ - 18, the sequence {BN} is an equal ratio sequence The first question has already been made, mainly the second one, which requires a specific process

The known sequence {an} and {BN} satisfy: A1 = λ, an + 1 = 2 / 3an + n-4, BN = (- 1) ^ n (an-3n + 21), where λ is a real number and N is a positive integer 1. Prove that the sequence {an} is not equal ratio sequence for any real number λ 2. Prove that when λ≠ - 18, the sequence {BN} is an equal ratio sequence The first question has already been made, mainly the second one, which requires a specific process

It seems that the second question prompts the first,
Let a (n + 1) = 2 / 3 * (an) + n-4 be
[A(n+1)-3(n+1)+21]=2/3[A(n)-3n+21]
So let C (n) = a (n) - 3N + 21
C1 = λ - 18, which is a sequence with a common ratio of 2 / 3
So B (n) is equal ratio, common ratio - 2 / 3
Given that the cube of X is equal to itself, find the value of x ^ 2 + 10x + 25 / (X-5) △ X / (x ^ 2-5x)
The cube of x equals itself, x = 1, x = - 1, x = 0
Substitute x ^ 2 + 10x + 25 / (X-5) △ X / (x ^ 2-5x) one by one
X=1 x^2+10x+25/(x-5)÷x/(x^2-5x)=36
x=-1 x^2+10x+25/(x-5)÷x/(x^2-5x)=16
X = 0, x ^ 2-5x = 0, so rounding off
The cube of x equals itself, x = 1, x = - 1, x = 0
Substitute x ^ 2 + 10x + 25 / (X-5) △ X / (x ^ 2-5x) one by one
X=1 x^2+10x+25/(x-5)÷x/(x^2-5x)=36
x=-1 x^2+10x+25/(x-5)÷x/(x^2-5x)=16
X = 0, x ^ 2-5x = 0, so no, so there are two answers: 16 and 36
The ellipse ax ^ 2 + by ^ 2 = 1 and the straight line x + y = 1 intersect at two points ab. the slope of the line between the midpoint C of AB and the center of the ellipse is √ 2 / 2. Find the slope of the ellipse
Let a (x1, Y1); B (X2, Y2) C (x0, Y0) 2x0 = X1 + x2; 2y0 = Y1 + y2a, B is the intersection of the ellipse ax ^ 2 + by ^ 2 = 1 and the straight line x + y = 1, ax1 ^ 2 + by1 ^ 2 = 1ax2 ^ 2 + BY2 ^ 2 = 1
Given that the lengths of two sides of a right triangle are 2 and 3 respectively, the length of the third pass is 2
According to the question, it is divided into the following three situations
(1) When two right angle sides are 2 (unit length) and 3 (unit length) respectively,
From the Pythagorean theorem, we have
The length of the third side is √ 13 (unit length)
(2) When 2 (unit length) is the third side, obviously, it does not hold
(3) When 3 (unit length) is the third side,
By Pythagorean theorem
The length of the third side is √ 5 (unit length)
13 or 5 under the root
The third side could be a root 5 or a root 13
The sequence {an}, {BN} satisfies A1 = k, a (n + 1) = (2 / 3) an + n-4, BN = (- 1) ^ n (an-3n + 21), where k is a real number and N belongs to n+
It is proved that the sequence {an} is not an equal ratio sequence. If {BN} is an equal ratio sequence, the range of K can be obtained
a(1)=k,
a(2)=(2/3)a(1)+1-4=2k/3-3=(2k-9)/3,
a(3)=(2/3)a(2)+2-4=(2/3)(2k-9)/3-2 = [4k-36]/9
If [a (2)] ^ 2 = a (1) a (3), then
(2k-9)^2/9=k(4k-36)/9,
(2k-9)^2=k(4k-36),
4k^2-36k+81=4k^2-36k,
81 = 0
Therefore, [a (2)] ^ 2 is not equal to a (1) a (3), {a (n)} is not an equal ratio sequence
a(n+1)=(2/3)a(n)+n-4,
a(n+1)+x(n+1)+y=(2/3)a(n)+n-4+xn+x+y=(2/3)a(n)+(x+1)n+x+y-4
=(2/3)[a(n)+3(x+1)/2*n + 3(x+y-4)/2]
x=3(x+1)/2,x=-3,
y=3(x+y-4)/2,y=12-3x=21.
a(n+1)-3(n+1)+21=(2/3)a(n)+n-4-3(n+1)+21=(2/3)a(n)-2n+14=(2/3)[a(n)-3n+21]
{a (n) - 3N + 21} is an equal ratio sequence with the first term of a (1) - 3 + 21 = K + 18 and the common ratio of 2 / 3
a(n)-3n+21=(k+18)(2/3)^(n-1)
b(n)=(-1)^n[a(n)-3n+21]=(-1)^n*(k+18)(2/3)^(n-1)=(-k-18)(-2/3)^(n-1)
-When k-18 is not equal to k-18,
{B (n)} is an equal ratio sequence with the first term (- k-18) and the common ratio (- 2 / 3)
1、
a(n+1)/an = 2/3 +(n-4)/an ①
a(n+2)/a(n+1) = 2/3 + (n-3)/a(n+1)②
If ① = ②
Then (n-4) / an = (n-3) / a (n + 1)
The solution is: an = 3 (n-3) (n-4) / (n-6);
a(n+1) = 3(n-2)(n-3)/(n-5) ≠(2/3)an+n-4
So {an} is not an equal ratio sequence.
2、
k≠19
What is the inverse function of F (x) = ln (5x-10)
It can be seen that y = in (5x-10) is symmetric with respect to y = x in time
That is: x = in (5y-10) and the exponent E
E ^ x = e ^ [in (5y-10)] according to the law of logarithm
e^x=5y-10
y=（e^x+10）/5
That's the answer
f（x）=（e^x）/5+2
The center of the ellipse is the origin o, the focus is on the x-axis, the eccentricity e = root 2 / 2, the line y = x + 1 intersects the ellipse at two points a and B, and the area of the triangle AOB = 2 / 3,
Find the equation of the ellipse
How to use calculator to calculate inverse trigonometric function,
Answer: old urchin
Senior
April 15, 18:26
tan20°+4sin20°=sin20°/cos20°+4sin20°
=(sin20°+4sin20°cos20°)/cos20°
=(sin20°+2sin40°)/cos20°=[sin20°+2sin(60°-20°)]/cos20°
=[sin20°+2(sin60°cos20°-cos60°sin20°)]/c0s20°
=[sin20+√3cos20°-sin20°]/cos20°
=√3
The sequence an is known to satisfy A1 = 2 and 3an + 1 = an, n = 1,2 ········, if BN = an + N, n = 1,2 ······,
Find the answer of the first n limits of BN and Sn
An + 1 / an = 1 / 3. {an} is an equal ratio sequence, so an = 2 (1 / 3) ^ n-1,
bn=an+n=2(1/3)^n-1+n
sn=2[1+1/3+(1/3)^2+----+(1/3)^n-1]-n+n(n+1)/2=3[1-(1/3)^n]+n(n-1)/2
Question: 3an + 1 = an
How to find the inverse function of function f (x) = x square - 5x?
If the domain of definition is (5 / 2, + infinity), there is an inverse function. The inverse function is: F-1 (x) = under the root sign (x + 25 / 4) + 5 / 2
First, f (x) = (X-5 / 2) ^ 2-25 / 4, and then write the inverse function according to the range of X. when x = 5 / 2, y = 5 / 2 + (x + 25 / 4) ^ 1 / 2
As for the known function is [- 25 / 4, ∞]
F (x) = (X-5 / 2) square-25 / 4
F-1 = (x + 25 / 4) root + 5 / 2, X ∈ [- 25 / 4, ∞]