In the sequence {an}, A1 = 1, a (n + 1) = 5 / 2-1 / an. Let BN = 1 / (An-2), find the general term formula of {BN} (2) let CN = - 3N * BN, find the sum of the first n terms of {cn}

In the sequence {an}, A1 = 1, a (n + 1) = 5 / 2-1 / an. Let BN = 1 / (An-2), find the general term formula of {BN} (2) let CN = - 3N * BN, find the sum of the first n terms of {cn}

(1) From BN = 1 / (An-2), an = 1 / BN + 2 a (n + 1) = 1 / b (n + 1) + 2
By introducing the relation between an and an + 1, we get 1 / b (n + 1) + 2 = 5 / 2-BN / (BN + 2), that is, B (n + 1 = 4bn + 2)
Then B (n + 1) + 2 / 3 = (B1 + 2 / 3) * 4, so BN + 2 / 3 is an equal ratio sequence BN + 2 / 3 = (B1 + 2 / 3) * 4 ^ (n-1)
Substituting the data, BN = - 1 / 3 * 4 ^ (n-1) - 2 / 3
(2)cn=n4^(n-1)+2n
Let DN = N4 ^ (n-1) kn = 2n
Then Sn = D1 + D1 +. DN, that is, Sn = 1 + 2 * 4 + (n-1) * 4 ^ (n-2) + n * 4 ^ (n-1)
Then 4sn = 4 + 2 * 4 ^ 2 + (n-1) * 4 ^ (n-1) + n * 4 ^ n
By subtracting one from the other, 3Sn = n4n ^ n - (4 ^ (n-1) + 4 ^ (n-2) +. + 4 + 1) = N4 ^ n - (4 ^ (n-1) - 1) / 3
So Sn = N4 ^ n / 3 - (4 ^ (n-1) - 1) / 9 reduces Sn = (4 ^ (n-1) * (12n-1) + 1) / 9
For kn SN1 = n * (n + 1)
So the sum of the first n terms of CN is SN2 = Sn + SN1 = (4 ^ (n-1) * (12n-1) + 1) / 9 + n * (n + 1)

a1=1 an=an_ 1+3n_ To the power of 1, n > = 2

Add up. That is an an_ 1=3^(n-1),∵a1=1,a2-a1=3,a3-a2=3^2,a4-a3=3^3…… an-an_ 1 = 3 ^ (n-1) ｜ the above equation is added on the left = added on the right, that is, an = 1 + 3 + 3 ^ 2 + 3 ^ 3 + +3^(n-1)=1×(1-3^n)/(1-3)=(3^n -1)/2

A1 = 3 A (n + 1) = an + 2 * 3N + 1 the general formula for finding an 3N represents the nth power a table of 3

a(n+1)-an=2*3^n+1
Therefore, there are:
a2-a1=2*3+1
a3-a2=2*3^2+1
...
an-a(n-1)=2*3^(n-1)+1
The sum of the above formulas is: an-a1 = 2 [3 + 3 ^ 2 +.. 3 ^ (n-1)] + (n-1) = 3 ^ n-3 + n-1 = 3 ^ n + n-4
So an = a1 + 3 ^ n + n-4 = 3 ^ n + n-1

The formula for finding BN by sequence an = (BN + 1) - BN B1 = 1 and an = 3n-2,
(BN + 1) is a number,

Bn+1-Bn=3n-2
Then BN = B1 + b2-b1 + b3-b2 + b4-b3 +. + bn-2-bn-3 + bn-1-bn-2 + bn-bn-1
=1+Sn-1
=1+(1+3n-2)(n-1)/2
=1+(3n²-3n-n+1)/2
=3n²/2-2n+3/2

B 1 = 1, B 2 = 2, B N + 1 * B N - 1 = B N to find the general formula of B n
I know it's a sequence with period six, but I won't prove it,

From B (n + 1) * B (n-1) = B (n), B (n + 2) * B (n) = B (n + 1)
From the above two formulas, B (n-1) * B (n + 2) = 1, that is, B (n) * B (n + 3) = 1, and then B (n + 3) * B (n + 6) = 1
From these two formulas, we get that B (n) = B (n + 6), that is, B (n) is a sequence of numbers with a period of 6

How to solve log2 (3x / x + 1) = - 1?

log2 (3x/x+1)=-1=log2 2^(-1)=log2 1/2
3x / x + 1 = 1 / 2, two sides multiply by 2 (x + 1)
6x=x+1
5x=1
x=1/5

The maximum value of function f (x) = log2 ^ (- 1 / 3x ^ 2 + 2x-1)

First, find the maximum value of (- 1 / 3x ^ 2 + 2x-1), he is the largest, the whole is the largest;
t=-1/3x^2+2x-1;
-b/2a=3;
Substituting Tmax = 2;
f(x)max=log22=1;
The maximum is 1;

log2（2x-1）＞log1/2（1-3x）

It can be seen from Log1 / 2 (1-3x) = - log2 (1-3x)
Log2 (2x-1) + log2 (1-3x) > 0
That is log2 ((2x-1) * (1-3x)) = log2 (5x-6x ^ 2-1) > 0
That is, 5x-6x ^ 2-1 > 1
6x^2-5x+21!

It is known that f (x) = log2 (x-1), G ((2 ^ x-t) / 2) = 2x (t ∈ R)
If it is true that G (x) is greater than or equal to f (x) when x belongs to 2 to 3, find the value range of real number t

Let x ＞ 1 and X & # 178; + 2x-11 = 0, then log2 (x-1) + log2 (x + 3) = (the simplest result must be written)

A:
x>1,x²+2x-11=0
x²+2x=11
log2(x-1)+log2(x+3)
=log2[(x-1)(x+3)]
=log2(x²+2x-3)
=log2(11-3)
=log2(8)
=log2(2³)
=3