For the problem of the mathematical equivalence sequence of equal differences, ask the master to ans

1. The common ratio is 1 2 The formula for summing is used in the proportional series.

2. sn=n(14n+6) 2 So d=14 a1=10 tn=n(2n+6) 2 d=2 b1=4

3、sn=1-1/2+1/2-1/3……1/n-1/(n+1)=1-1/(n+1)

4、sn=(1+2+4……2^n)-n+(1+2n-3)n/2

Landlord classmate: These questions are very basic, and there are formulas in the book, just substitute them.

You have to read a book, otherwise you won't be able to keep up in the future, I hope you listen to me.

1 The formula for the infinity of the sum of this series is a (1-r).

a is 1 and the first number r is 1 2 are equal to each number.

Isn't it very simple to substitute 2 and you slowly substitute formulas?

So when these two sequences are combined, what kind of socks question will be generated? In this issue, I will bring you a few such questions.

Let's take a look at the following question.

Although this is a proportional series, a concept called the equivalence term is used.

Using the properties of the proportional series, all terms are highlighted by the table a2 and q, and the two sides of the equal sign are reduced to a2 at the same time to obtain a quadratic equation about q.

Solve this equation, and because each item is a positive number, round off the negative values, and the final answer is obtained.

Difference and proportionality are two special sequences that can be converted into each other by taking logarithms or exponential powers. Therefore, sometimes the problem of proportional sequences will be examined in combination with the nature of logarithmic operations, such as the following question.

The logarithms of the same base are added, the base number does not change, and the true number is multiplied.

According to the nature of the proportional terms, the product of the first five terms is only related to the third term. Finally, combined with the logarithmic arithmetic, the final answer can be obtained.

Finally, let's take a look at one such question, which is the 2021 final exam question in Suqian, Jiangsu.

We need to first find the general term formula for the series based on the known conditions.

Finally, an is converted into a base exponential power of 2, so that we can further observe how to do it next.

What we require is the maximum value of the product of the first n terms of the series, an is the exponential power with 2 as the base, and multiply it with the power of the base, and the invariant exponents of the base are added, and finally the macroization is conveyed into the maximum value of the sum of the first n terms of the equal difference series.

How do you arrive at this series of equal differences? It's as simple as taking the logarithm of 2 for an.

Let's see how you have mastered the content of the last issue, do you remember the two methods of finding the first n terms and the most value of the equal difference series? Here we use the method of quadratic functions, first find the first n terms and sn

Then, by judging the opening direction and the axis of symmetry, the maximum value of SN can be calculated. Note that n can be a positive integer.

Finally, let the product of the first n terms of the series be tn, and obtain the relationship between tn and sn, and then the maximum value of tn can be obtained from the maximum value of sn.

(1) Observational induction.

This method requires strong reflexes!

For example, 21, 203, 2005, 20007 can you quickly see this?

2) Accumulation and commercial method (we call it superposition and superposition in our textbooks, and I won't say much about the specific books).

For example, a1 is known, and a(n+1)-an=f(n).

a1 is known, and a(n+1) an=f(n).

3) Constructive method.

This method is the most difficult, but once you master the technique, you will be able to solve any problem.

For example, if we know a1, the form of a(n+1)=pan+q can be constructed, that is, it is matched as a(n+1)+x=p(an+x) Of course, the middle minus sign is the same!

For example, the sequence satisfies a1=1, a(n+1)=1 2 an+1

Solution: Let a(n+1)+a=1 2(an+a) and then put a zero pending coefficient, this item should be equal to the item of the original problem!

4) Formula method.

I don't need to talk about this method! Two formulas, equal difference, equal ratio! Not using the question is often not as simple as testing you, and often set a trap, maybe n=1 is often not taken into account! So be cautious when doing the questions!

Let the tolerance of the equal difference series be d, and the common ratio of the equal ratio series be q

a2+b2=a1+d+b1×q=1+d+3q=8

d=7-3q

t3-s3b1+b2+b3-a1-a2-a3

b1(1+q+q^2)-(3a1+3d)

3(1+q+q^2)-(3+3d)

15q+q^2-d=5

q+q^2-7+3q=5

q^2+4q-12=0

q+6)(q-2)=0

q=-6 (rounded, the common ratio is positive) or q=2

So q=2d=7-3q=7-3*2=1

an=a1+(n-1)d

1+(n-1)*1

nbn=b1q^(n-1)

3*2^(n-1)

cn+2c(n-1)+3c(n-2)+4c(n-3)+.n-1)c2+nc1

cn+2c(n-1)+3c(n-2)+4c(n-3)+.n-1)c2+nc1=2^(n+1)-n-2 ..1

c(n-1)+2c(n-2)+3c(n-3)+.n-2)c2+(n-1)c1=2^n-(n-1)-2(n>=2) .2

1-2 formula.

cn+c(n-1)+c(n-2)+.c3+c2+c1=2^n-1(n>=2)..3

So c(n-1)+c(n-2)+c3+c2+c1=2^(n-1)-1(n>=3) .4

Type 3-4.

cn=2^(n-1)(n>=3)

When n=1,2, the above equation is suitable.

So cn=2 (n-1).

That is, the number series is a proportional number series.

s3=a1+a2+a3=3a2（a1+a3=2a2）

t3=b1+b2+b3=3+b2+b3

a2=8-b2

t3-s3=3+b2+b3-3a2=3+b2+b3-24+3b2=15

4b2+b3-36=0

4b1q+b1q^2-36=0

q^2+4q-12=0

q+6)(q-2)=0

q=-6 q=2

b2=6 b2=-18

a2=2 a2=26

an=n bn=3*2 (n-1) or an=25n-24 bn=3*(-6) (n-1).

I brought an=n as an example to prove the second question, the same is true in the other case, too troublesome not to write.

It is known that 1cn+2c(n-1)+3c(n-2)+4c(n-3)+n-1)c2+nc1=2^(n+1)-n-2

Bringing in n=n-1 has 1c(n-1)+2c(n-2)+3c(n-3)+n-2)c2+(n-1)c1=2^n-n-1

Subtract the above formula to obtain cn+c(n-1)+c(n-2)+c(n-3)+c2+c1=2^(n+1)-2^n+1=2^n-1

cn+c(n-1)+c(n-2)+c(n-3)+.c2+c1=2^n-1

Bring in n=n-1 to get c(n-1)+c(n-2)+c(n-3)+c2+c1=2^(n-1)-1

Subtracting the above formula yields cn=2 n-2 (n-1)=2 (n-1).

So it turns out that it is the ratio of c1=1 q=2.

In the equal difference series an, it is known that a4+a7+a11+a14=128, a6+a9+a12=96

If s10 is greater than 0 and s11 is less than 0, then the minimum n value that makes an less than 0 is 6

In the proportional series, (1) it is known that a1+a2+a3=5, a4+a5+a6=135, a8=5*3 7 13

2) a4, a12 are the two roots of the equation 2x*x-21x+8=0, a8=2

3)a6+a5=a7-a5=48,s10=__1023___

In fact, it is very simple to prove whether it is equal or equal difference, that is, to grasp the definition of the equal difference series, that is, an-a(n-1)=d, that is, the subtraction is equal to a constant, and the proportional series is an a(n-1)=q, that is, after division, it is equal to a constant, if it is a compound type, it may be more troublesome.

For example, if you ask the same term of an in an=2a(n-1)-1, as long as you set an x, it is OK, if you encounter this kind of thing, it is a proportional series after compounding.

Combined with the above, the proportional sequence is that the adjacent two terms are divided as constants, because to construct the adjacent terms, let an+x=2*(a(n-1)+x) and then disassemble, because the left side of the original series is -1, so move x to the left, 2x-x=-1 to calculate x=-1 and then substitute it into the original series.

an-1=2*(a(n-1)-1), so an-1 is a proportional series of 2.

Then you use the general term formula of the proportional sequence to find the formula of an-1, and then -1 is OK, if you encounter an-2a(n-1)=2 (n-1), just divide the two sides by a 2 n (that is, the maximum number of terms), and then use my method above to continue. That's it for today.

It is only necessary to prove that f(n+1)=a*f(n) can show that fn) is a proportional series with a as the common proportion, and the key is deformation, such as using s(n+1)-s(n) instead of a(n) and more practice, accumulation of skills.

1.Find out the general term, 2Compare two adjacent items, and the result is constant.