Mathematics questions: 25 for the number series questions first, and 15 can be given after answering

1: Proof: an+1-an)g(an)+f(an)=0

an+1-an)*10(an -1)+(an -1) 2=0 gives an -1=-10(an+1 -an).

an-an-1)*10(an-1 -1)+(an-1 -1) 2=0 gives an-1 -1=-10(an -an-1).

Yes: (an -1) (an-1 -1) = (an+1 -an) (an -an-1).

Simplify: (an -1) 2=(an+1 -1)(an-1 -1) From the above, it is known that none of the terms of this equation is zero.

Therefore, the available sequence is an equal proportional series.

2:t^m/b^m＜t^(m+1)/b^(m+1)

t/b)^m<(t/b)^(m+1)

Take the logarithm on both sides. To simplify: ln(t b)>0 b is always positive, then t is also always positive.

Obtain: t>b=max(bn).

bn= -1)=

bn>bn-1 is incremental.

n+2) >n+1)

When n<8, bn is incremental.

bn>bn+1 is decreasing.

n+2)>(n+3)^

When n>7, bn is decreasing.

Substituting n = 7 and 8, there is b7 = b8 = 10*

Therefore, the maximum value of bn is 10*

t>10*

Solution: Substitute f(x) and g(x) first.

Yes, (an+1-an)*10*(an-1)+(an-1) 2=0an-1)(10an+1-9an-1)=0, then an-1=0 or 10an+1=9an+1a1=2 10an+1=9an+1

10(an+1-1)=9(an-1)

an+1-1)/(an-1)=9/10

That is, a proportional series with 1 as the first term and the common ratio of 9 to 10.

It's not that everyone doesn't do it for you, you have a problem with this question, or you don't give all the conditions, or you make a mistake.

4, 24, 6, 36, 8, 54, this sequence.

All odd digits are 4, 6, 8, and the shroud.

Then the following digits are 10 and 12

The even digits are 24, 36, and 54

They are 6 times, 6 times, and 7 times the odd digits, so we can see that the following even digits should be 70 and 96.

So 10, 70, 12, 96 should be filled in the following parentheses.

1. To learn the number series, you must first master some basic formula points. For example: find the general term, find the sum of the first n terms;

2. You should remember the basic sequence formula, after all, the formula is like a brick for building a wall, you can't build a wall without bricks, on this basis, take a look at the example problems, the example problems must be representative;

3. Familiarize yourself with the formula by learning more and doing more;

4. Understand the question types of number series, such as: abstract number series question types, combined with functions;

5. After that, try to do simple questions;

6. Slowly increase the difficulty, so that it will be easier to master and learn to solve the number series.

To learn the number sequence, you first need to make some basic announcements. To find the general term, find the sum of the first n terms, several formulas are definitely required, such as s(n+1)-sn=an This is the most basic way to find the general term.

Mathematics is about learning more and doing more. The formula is not for memorization, it is cooked...

The question type of the number series in the college entrance examination is not difficult, it is nothing more than finding the middle term of an, sn and other ratios and the middle term of equal difference are often used.

Here's my personal opinion. There are three levels of the number series in high school.

1) Basic question types. You need to summarize this yourself, such as multiplication and common ratio, dislocation subtraction, list method, and so on. These things feel difficult to learn at first, and you will definitely not be used to them, so you need to summarize the question types and then practice according to the question types.

After a while, you will be able to get full marks for this section.

2) Think about the question type. This section often has some regular questions for you to find out the general formula, or to overturn it. The specific methods are ever-changing, and there is no good response to it, only one can be encountered and one can be killed.

3) Puzzles, don't think about this part, use to put things that these universities won't ask for... The college entrance examination is generally the last finale question. . . And it's the level of the third question.,Let's not want it.,Just take the test 144.。。。

In order to get a high score, you have to focus on part (1). Part (2) After passing part (1), do the questions appropriately. Part (3) suggests abandonment, which makes no sense.

Site area = 40x26 = 1040m

Grassland area = 144x6 = 864m

Road area = site area - grassland area = 1040-864 = 176m Set the road width = xm, road area = 40x+2 26x-2x (intersection double count, to be deducted).

92x-2x²=176

46x-x²=88

x²-46x+88=0

x-2）（x-44）=0

x=2, or x=44 (impossible).

The width of the road is 2m

On your own, no one is a nano scoundrel oh go online to answer your questions, and you can't see clearly.

The brigade was aggressively annihilated and dismantled, 94 + 9 + 4 = 107

Divided by 3 by 1 and 5 by 1, it means that the difference between the neighbors of this number is 15, and the number is satisfied, so the direction of this number is 15n-14, then 15n-14=2016, the solution is n=135, and n is an integer.