It is known that x, f x 2, 3 x 0 are equal difference sequences

I'm not going to help you make it, let's talk about the solution idea:

An equation can be made according to the first sentence:

x-[√f(x)]/2=[√f(x)]/2-√3,..1) According to the second sentence, another equation can be listed:

s1=a1=3;

s2=a1+a2=3+a2=f(s1)=f(3)..2) Substituting x=3 into equation (1) yields:

3-[√f(3)]/2=[√f(3)]/2-√3...3) The specific value of f(3) can be found according to equation (3).

Find f(3) and then substitute (2) to get the specific value of a2.

Alas, let's not talk about the rest of the story, anyway, that's the way of thinking.

Solution: x, f(x) 2, 3 into a series of equal differences, then.

2√f(x) /2=√x+√3

f(x)=√x+√3

f(x)=x+2√(3x) +3

N2, sn=s(n-1)+2 [3s(n-1)].

sn-s(n-1)-3=2√[3s(n-1)]

an-3=2√[3s(n-1)]

12s(n-1)=(an-3)²

12sn=[a(n+1)-3]²

12sn-12s(n-1)=12an=[a(n+1)-3]²-an-3]²=a(n+1)²-6a(n+1)-an²+6an

a(n+1)²-an²-6a(n+1)-6an=0

a(n+1)+an][a(n+1)-an]-6[a(n+1)+an]=0

a(n+1)+an][a(n+1)-an-6]=0

an>0, a(n+1)+anconstant》0, to be the equation true, only a(n+1)-an=6, is a fixed value.

s2=s1+2√(3s1) +3

a1+a2=a1+2√(3a1)+3

a2=3+2√3

From the second term, the series is a series of equal differences with 3+2 3 as the first term and 6 as the tolerance.

When n 2, a(n+1)=a2+6(n-1)=6n+2 3

3, n=1, a2=3+2 3, also satisfied.

a(n+1)=6n+2√3 -3

bn is a proportional sequence of 1 a(n+1), 1 an, then.

bn)²=[1/a(n+1)][1/an]

bn=1/[ana(n+1)]

When n=1, b1=1 (a1a2)=1 [1 (3+2 3)]=3-2 3

When n 2, bn = 1 [ana(n+1)]=1 [(6n+2 3-9)(6n+2 3-3)]=(1 6)n=1, t1=b1=3-2 3

At n 2, tn=b1+b2+.bn

1/[6n+2√3-3]]

3-2√3+(1/6)[(3-2√3)-1/[6n+2√3-3]]

7/6)(3-2√3) -1/[6(6n+2√3-3)]

1) In the equal difference series {an}, an=a1+(n-1)d (d is the tolerance) a2=a1+d, a3=a1+2d

a1=2,s3=26

Funny Qing a2 + a3 = 2a1 + 3d = s3-a1 = 24d = 20 3

Q made a mistake).

2) In the proportional sequence ruler{an}, an=a1*q (n-1), sn=(a1-an*q) 1-q

a1=1/2,an=243/2 ,sn=182(1/2- 243/2 *q)/(1-q)=182q=3an=1/2*3^(n-1)

Let an=243 2 =1 2*3 (n-1) then n=6q=3,n=6

If you're satisfied, remember!

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Hee-hee......I'm drinking Coca-Cola in the desert, singing karaoke, riding a lion and chasing ants, and answering mountain questions for you with a keyboard in my hand!

Difference series: a2-a1=a3-a2

a2=0 so:

a1+a3=0

Substituting f(x+1) and f(x-1) into a1+a3=0, there is:

x 2-4x+3=0, so when x=1 or x=3x=1, substitute a1=f(x+1) to get a1=-2, a2-a1=2=d, and an=-2+2(n-1).

When x=3, substituting a1=f(x+1) gives a1=2, a2-a1=-2=d, and an=2-2(n-1).

f(x+1)=x^2-4=(x+1)^2-2(x+1)-3f(x)=x^2-2x-3

f(x-1)=(x-1) 2-2(x-1)-3=x 2-4x equal difference series. a1+a3=2a2

x^2-4x+x^2-2x-3=-3

2x^2-6x=0

The solution is x=0 or x=3, and a1=0, a3=-3, so an=-3(n-1) 2

The third question is that you can do it yourself, and every time the difference series is extracted from the fixed value, it is still the difference.

an=2n-4；sn=n^2-3

an=-2n+4;sn=-n^2+3

I haven't touched this thing in years. Probably not

1) x=-1 is the common root, which can be obtained by observing it, or by substituting the general term of the number series.

2) What are you looking for?Can you make it clearer?

1+4+7+..x=(1+x)[(x-1) 3+1] 2=590, so the square of x 3-1 3+1+x=1180x2+3x-3538=0

Use a quadratic equation to find the root formula: x=58 or -61 (negative values are rounded), so x=58

This is the junior high school method.

If you're an elementary school student, use the upstairs method).

Let x=3n-2 be summed with equal difference to get n=20, then x=58

Because x>3

So only the front burn X-3>0

Therefore, the original formula = 4 + x(x-3) >7(x-3) is arranged to get the penitent tomb (x-5) 2 >=0

Therefore, the original form is a false one.