Guys, prawns, help me do the high school math problem 15

an＋2 -an＋1 ＝（2/3）an＋1－（2/3）an

a2 -a1= 2/3

So. an＋2 - an＋1＝（2/3）^n

an＋1 - an＝（2/3）^（n-1）

an - an-1＝（2/3）^(n-2)

a2 - a1＝2/3

a1 = 1 above the equation accumulates, a(n2 ) = 3-2(2 3) n

an= 3-2(2/3)^（n-2）

sn= 3n- 2×《 1×(2/3）^(1) +2×(2/3）^(0) +3×(2/3）^(1)……n×(2/3）^(n-2)》

Let tn =1 (2 3) (1) +2 (2 3) (0) +3 (2 3) (1)......n×(2/3）^(n-2)

2/3tn = 1×(2/3）^(0) +2×(2/3）^(1)……n-1×(2/3）^(n-2)+n×(2/3）^(n-1)

Subtract the two formulas, 1 3 tn = 7 2-(n+2)(2 3) (n-1).

sn= 3n- 2× tn=3n- 21+ 6（n+2)(2/3）^(n-1)

a(n+2)-a(n+1)=2/3(a(n+1)-a(n))=>a(n+2)-a(n+1)=(a(2)-a(1))*2/3)^(n-1)=(2/3)^n

a(n+2)=(a(n+2)-a(n+1))+a(n+1)-a(n))+a(2)-a(1)）+a(1)

2/3)^n+(2/3)^(n-1)+…2/3)^0+1=……nan=……

You can calculate it yourself later.

The eigenroots are 2 3 and 1, and then an=-3(2 3) n+3 is obtained

Then count the sum of the first n terms of the nan.

I finally understand, I'll prompt it, divide (5 3)an into an + (2 3)an, and move it over.

Upstairs is amazing!

Summary. Solve a high school arithmetic.

2 x 2x

x(x-1)^2（x^2-4)=x(x-1)(x-1)(x+2)(x-2)<0

Then use the sequence shaft stubbing method:

The answer is x<-2, 00

First, find the root x=0,x=1,x=-2,x=2,x=3 of the equation x(x+2)(x-1)(x-3)=0

Highlight the points one after the other on the number line. Or a curve is drawn from the upper right of the rightmost point 3, passing through point 3, resembling an opening upwards parabola between , passing point 1; Continuing to the upper left of point 1, this curve resembles a curve with an opening downward between the points, passing through point 0; Continue to the lower left of 0, between 0 and -2, similar to a parabola with an opening upward, passing through the point -2; Continue to extend infinitely to the upper left of point-2.

The requirement in the equation is 0

Just look at the range of x taken by the part of the curve above the number line.

x<-2 or 03

where (x-1) 2>=0 is constant, i.e., x is not equal to 0, then (x-1) 2 terms Evergrande is 0

The original problem is equivalent to the solution set of (x+2)(x)(x-2)<0, and x is not equal to 0, then the solution set of the original inequality is 0< x<2 or x<-2< p>

(x-1) When 2 x is not equal to 1, this factor greater than 0 gives a solution x is not equal to 1

The inequality is x(x 2-4)<0

x(x+2)(x-2)<0

x(x+2)(x-2)=0 is solved to obtain x=0 x=2 x=-2 inequality solution as x<-2 0 inequality solution set x<-2 0< x<2 x is not equal to 1< p>

x is not equal to 1 or plus or minus 2 or 0

x-1)^2>0

So x(x2-4)=x(x+2)(x-2)<0, then x<-2 or 0

Use the threading method.

infinity, -2) or (0,1) or (1,2).

Is that (x-1) 2 the square of (x-1)? If yes, you look at the big case, isn't it: 2< x<0 and 1

When a=1 is, it holds.

When a is not = 1, the coefficient of the second term is greater than 0 and greater than 0

It would be nice to solve it.

This kind of topic needs to be discussed.

The first case.

a squared - 3a + 2 = 0

a=1 or a=2

When a=1, the above equation = 2>0, for any x constant when a=2, the above equation = x+2, for any x is not constant" 0, so 2 rounds off the second case, a squared -3a + 2>0

b 2-4ac=(a-1) 2-4(a 2-3a+2)*2<0 to obtain a 1 or a 15 7

Hence the intersection with the first case.

The solution is less than or equal to 1 or a 15 7

Hope it helps, by the way, the one upstairs forgot about the case where the quadratic coefficient is 0.

It's been done a long time ago! The answer is ———x!

(a's square - 3a+2) >0

b 2-4ac 0 (formula ha, bring yourself).

Just solve it.

Consider when the quadratic coefficient is zero.

(1) The secondary term is not 0

Drawing: Because (a -3a+2) x + (a-1) x + 2 0 is true for x, so: the opening is upward, i.e. a -3a+2 0

There is no intersection with the x-axis, i.e. 0

Calculated: a 1 or a 15 7

b) The quadratic term is 0 (i.e., a = 1 or 2).

Substituting it can be seen that a=1,2 0 is true.

a=2, not true.

So to sum up, a 1 or a 15 7

(a -3a+2)x +(a-1)x+2>0 is established on r (1)a -3a+2>0 only when the opening is upward can the question (2) =(a-1) -4(a -3a+2)2<0 be comprehensively solved to obtain x (-1) (15 7,+

a< = 1 or a>17 7

When a 2-3a+2=0, a-1=0 is required, and the solution is a=1;

When a2-3a+2 is not equal to 0.

a^2-3a+2>0

Discriminant formula (a-1) 2-4*2*(a 2-3a+2)<0 solution to a<1 or a>17 7;

In summary, a<=1 or a>17 7.