Let a[n] be a geometric series.

A[n] cannot be all zeros. This is because if a[ 1]=x=0, then a[2]=-3, which is not zero.

Therefore, all a[n] must be non-zero.

a[ 1]=x，a[2]=2x/3-3，a[3]=2/3*(2x/3-3)-2=4x/9-4

If a[n] is equal, then a[2]/a[ 1]=a[3]/a[2]:

2/3-3/x=(4x/9-4)/(2x/3-3)

Multiply both sides by x: 2x/3-3 = (4x/9-4)/(2x/3-3) * x.

Multiply both sides by (2x/3-3): (2x/3-3) 2 = (4x/9-4) * X.

9=0, no solution. Therefore, for any number x, a[n] cannot be a geometric series.

(2)

First, find the general formula of a[n]:

a[n+ 1]=(2/3)a[n]+n-4

A[n]=(2/3)a[n- 1]+n-5, that is, (2/3) a [n] = (2/3) 2 * a [n-1]+(2/3) * (n

……

A[2]=(2/3)a[ 1]-3, that is, (2/3) (n-1) A [2] = (2/3) n * x-3 * (2/3) (n-.

Add up all the formulas to get:

A [n+1] = (2/3) n * x+σ (I =1to n) ((I-4) * (2/3) (n-I))

= (2/3) n * x-4σ (I = 1 to n) ((2/3) (n-I))+σ (I = 1 to n) (I * (2/3) (n-I))

= (2/3) n * x-4σ (I = 0 to n-1) ((2/3) i)+((2/3) (n-1)+2 * (2/3) (n-2)

=(2/3)^n*x-4( 1-(2/3)^n)/( 1-2/3)+((2/3)^0+((2/3)^0+(2/3)^ 1)+……+((2/3)^0+……+(2/3)^(n- 1)))

=(2/3)^n*x- 12( 1-(2/3)^n)+(( 1-(2/3)^ 1)/( 1-2/3)+( 1-(2/3)^2)/( 1-2/3)+……+( 1-(2/3)^n)/( 1-2/3))

=(2/3)^n*(x+ 12)- 12+3*(( 1-(2/3)^ 1)+( 1-(2/3)^2)+……+( 1-(2/3)^n))

=(2/3)^n*(x+ 12)- 12+3*n-3*((2/3)^ 1+(2/3)^2+……+(2/3)^n)

=(2/3)^n*(x+ 12)- 12+3*n-3*(2/3)*( 1-(2/3)^n)/( 1-2/3)

=(2/3)^n*(x+ 18)- 18+3*n

Therefore, b [n] = (-1) n * (a [n]-3n+21) = (-1) n * (2/3) n * (x+18).

=(-2/3)^n*(x+ 18)+3*(- 1)^n

s[n]=(x+ 18)*(-2/3)( 1-(-2/3)^n)/( 1-(-2/3))+3*(- 1)*( 1-(- 1)^n)/( 1-(- 1))

=(x+ 18)*(-2/5)( 1+(- 1)^(n+ 1)*(2/3)^n)-3/2+(3/2)*(- 1)^n

=(-(2/5)*(x+ 18)-3/2)+(- 1)^n*(3/2+(2/5)*(x+ 18)*(2/3)^n)

According to the expression of S[n], the first term has nothing to do with n, and (3/2+(2/5) * (x+ 18) * (2/3) n in the second term is monotonous (not sure whether it is increasing or decreasing). Therefore, when n= 1 or 2 or n tends to infinity, the maximum and minimum values must be taken.

s[ 1]=-3-(2/3)*(x+ 18)

S[2]=(-2/9)*(x+ 18)

When n tends to infinity, when n is odd or even, S[n] tends to:

Odd number: (-2/5)*(x+ 18)-3

Even number: (-2/5)*(x+ 18)

These four formulas are required to be in the interval (a, b) respectively. I don't understand inequality.