High school math conjecture positive number series general term formula

Convention: [ inside is the subscript a[1]=(1 2)(a[1]+1 a[1]) solution a[1]=1

Solution: n=1: s[1]=1 (s[1]) 2=1 when n>1.

2s[n]=(s[n]-s[n-1])+1/(s[n]-s[n-1])

s[n]-s[n-1]=1/(s[n]-s[n-1])s[n])^2-(s[n-1])^2=1

So it is a series of equal differences where the first term is 1 and the tolerance is also 1.

(s[n]) 2=n

and s[n]>0 has s[n]= n

a[1]=1=(√1)-(1-1))

When n>1 a[n]=s[n]-s[n-1]=( n)- n-1), so a[n]=( n)- n-1).

Hope it helps!

The method of finding the formula for the general term of the number series:

1. The first n terms and sns are known

Exploit to solve.

2. Known recursive relationship.

Using the undetermined coefficient method, a new series (equal ratio or equal difference) is obtained, and the general term formula of the new series is obtained by using the summation formula, so as to solve the general term formula of the original number series.

Other methods: find the period of the series, take the reciprocal, commutation method (touching the root number), iterative and superposition methods, ......

The formula for the first n terms of the high school mathematics number series is sn=n*a1+n(n-1)d 2 and the first n terms of the equal difference series: sn=n*a1+n(n-1)d 2 or sn=n(a1+an) 2. The general formula for the equal difference series is:

an=a1+(n-1)d。

If a series of numbers is erected from the second term, and the difference between each term and its previous term is equal to the same constant, this series is called the equal difference series, and this constant is called the tolerance of the equal difference series, and the tolerance is often represented by the letter D.

There is a formula for summing the difference series.

The equation of the difference series is an=a1+(n-1)d,

The sum of the first n terms is: sn=na1+n(n-1)d 2, if the tolerance d=1: sn=(a1+an)n 2, if m+n=p+q:

If am+an=ap+aq, if m+n=2p, then am+an=2ap, all of the above n are positive integers.

The formula of the general term of the number series is the important and difficult point of high school mathematics, so what are the solutions to the general term formula of the number series? Let me tell you the answer.

1. The problem of finding the general term of the first-order linear recursive sequence.

There are several main forms of first-order linear recursive sequences:

This type of recursive sequence can be obtained by accumulation to find its general formula (the series can find the sum of the first n terms).

While. When it is a constant, the general formula of the equal difference series can be obtained by the accumulation method. And when.

If it is a series of equal differences, then.

is a second-order equal difference series, and its general formula should be .

form, note the difference from the general form of the equation for summation of equal difference sequences, the latter is.

Its constant term must be 0 2.

This type of recursive sequence can be obtained by multiplication to find its general formula (the product of the first n terms can be found for the series).

While. When it is a constant, the general formula of the proportional series can be obtained by multiplication. 3.

Such sequences can usually be converted to.

or eliminate the constant and convert it to second-order recursion.

Example 1: A known sequence.

Medium, seeking. The general term formula. Analysis: Solution 1: Transform into.

type recursive series. ∵

Again. Therefore, the number series satisfies a1=1 2, a(n+1)=an+1 (4n 2-1), and the formula for the general term is solved.

Solution: a(n+1)=an+1 (4n 2-1)=an+[1 (2n-1)-1 (2n+1)] 2

an=a1+(1-1/3+1/3-1/5+……1/(2n-3)-1/(2n-1))

an=1/2+1/2 (1-1/(2n-1))=4n-3)/(4n-2)

Multiplication. The recursive formula is a(n+1) an=f(n), and f(n) can be quadratified.

For example, if the sequence satisfies a(n+1)=(n+2) n an, and a1=4, find an

Solution: an a1=an a(n-1) a(n-1) a(n-2) ....a2/a1=2n(n+1)

Construct. Convert non-proportional series and proportional series into related proportional series.

Add and subtract, and multiply and divide.

Example: A1 + 2 A2 + 3 A3 + ......nan=n(n+1)(n+2)

Solution: Let bn = a1 + 2a2 + 3a3 + ......nan=n(n+1)(n+2)

nan=bn-b(n-1)=n(n+1)(n+2)-(n-1)n(n+1)

<> this topic is relatively simple, that is, the basic method of finding the number series.

Enumerate it and add the left side to the left, and the right side is very right, and you can eliminate a lot of phases.

The specific process is written on paper

List, add the two sides, and calculate as follows.

The content comes from the user: September love in the world is strong.

The method of finding the formula for the general term of the number series:

1. The method of finding the general term formula of the number series that needs to be mastered: the observation induction method, the formula method, and the general term formula of the known number series. What needs to be mastered is to be extremely proficient in using it and can be completed at any time.

1.Observational induction:

Example 1: According to the values of the first few terms of each series of numbers, write a general formula for the following series.

Analysis: (1) It is not difficult to guess from ,,,:.

2) Each term of the sequence can be reduced to a fraction, so it should be studied in terms of both the numerator and the denominator. The characteristics of the molecule are more obvious, so it can be seen that it ,,, conjecture.

Note: A general formula for conjecturing a series of numbers from the values of the first few terms of the series adopts the incomplete induction method, and the result obtained may be wrong, and the mathematical induction method is used to prove it in the solution problem. But it's a good way to do it in small questions like multiple-choice and fill-in-the-blank questions.

If the known sequence satisfies, then =().

a 0b c d solution: the recursive formula is known, let , in turn,,, it is not difficult to guess that the number series is a special series with a period of 3, therefore, choose b.

2.Formula Method:

Example 2: In a known sequence, the point is on a straight line, and the general formula of the sequence is found.

Analysis: Derived from the title: , ie.

The sequence is a first-term, equal-difference series with a tolerance of 7. ∴

Summary: Through appropriate transformation of the topic, the number series conforms to the definition of equal difference series and proportional series, so as to use the general formula of equal difference series and proportional number series to solve.

3.The general formula for finding a series of numbers is known:

Example 3: Knowing the sum of the previous terms of the following series, the general formula is found.

1 Analysis:

f(n+2) = f(n+1) +f(n) => f(n+2) -f(n+1) -f(n) = 0

Let f(n+2) -af(n+1) = b(f(n+1) -af(n)).

f(n+2) -a+b)f(n+1) +abf(n) = 0

Obviously a+b=1 ab=-1

From Veda's theorem, we know that a and b are the two roots of the quadratic equation x 2 - x - 1 = 0.

The solution gives a = (1 + 5) 2, b = (1 - 5) 2 or a = (1 - 5) 2, b = (1 + 5) 2

Let g(n) = f(n+1) -af(n), then g(n+1) = bg(n), and g(1) = f(2) -af(1) = 1 - a = b, so g(n) is an equal proportional series, g(n) = b n, i.e.

f(n+1) -af(n) = g(n) = b^n --1)

In equation (1), the above two sets of solutions of a b are substituted respectively, and because of the symmetry, x = (1 + 5) 2 and y = (1 - 5) 2 are set to obtain:

f(n+1) -xf(n) = y^n

f(n+1) -yf(n) = x^n

Subtract the above two formulas to get:

x-y)f(n) = x^n - y^n

f(n) = (x^n - y^n)/(x-y) = /√5

Look at this page, this is actually a Fibonacci sequence.

1. The definition of the general term formula of the number series: a series of numbers arranged in a certain order is called a number series, and the nth term of the number series is expressed by a specific formula (containing parameter n), which is called the general term formula of the number series. This is like the analytic expression of a function, which can be found by substituting the specific n value to find the value of the corresponding an term.

The method of finding the general term formula of the number series is usually obtained by the recursive formula through several transformations.

2. There are two series of equations of equal difference in the first year of high school.

and the general term formula for proportional series.

If the tolerance is d, then an=a1+(n-1)d, which is the general formula for the series of equal differences.

Note: 1) Because an=nd+(a1-d), Hand Mo Hui so the image of the equal difference series is a natural number in the abscissa.

Some scattered points on the same straight line of a column, the geometric meaning of tolerance d is the slope of that line.

2) The general formula for the difference series can also be determined by the following formula: an=am+(n-m)d, am+n=(mam-nan) (m-n).

3) The tolerance d of the equal difference series can be determined by the formula d=(an-am) (n-m).

If the first term of the proportional series is a1 and the common ratio is q, then the general term formula for the series an is an=a1qn-1

Note: 1) Because an=a1qn-1, when q>0 and q≠1, the image of the proportional series is the same exponential function with the abscissa of the natural number.

on some scattered points.

2) The general formula of the proportional series can also be answered by the formula an=amqn-m.

Example 4 In a known proportional series, a1=1 and a2=2, write the general formula.

3. There are many ways to find the general term formula and the real number series, for example, the direct method and the formula method.

Inductive guessing ideas, accumulation, multiplication, reciprocal, logarithmic.

Iterative method, pending coefficient method.

Fixed point method, commutation method, periodic series, eigenroot method, ......Wait a minute!

a(n+1)=2an (2+an).

Take the bottom of both sides.

1/a(n+1)=1/an+1/2

1/a1=1

1 an} is a series of silver rent equal difference width shouting with 1 as the first term and 1 2 as the tolerance.

1/an=1+(1/2)*(n-1)

1/an=(n+1)/2

an=2/(n+1)

Accumulation: a2-a1=3

a3-a2=3^2

a4-a3=3^3

an-an-1=3^n-1

Accumulate an-a1=[3(1-3 n-1)] -2=(3 n-3) 2 (n 2).

an=(3^n-3)/2+1（n≥2）

When n=1, a1=0+1=1 satisfies.

an=(3^n-3)/2+1

This is a series of numbers that are proportional to the difference!! I first think of a(n+1)-an as a new book column bn, then the sum of the first (n-1) terms of bn is (3 2)(3 (n-1)-1).Then an=(3 2)(3 (n-1)-1)+1

You can use the accumulation method a2-a1=3 a3-a2=3 2 a4-a3=3 3··· an-an-1=3 n-1 all plus so an-a1=3+3 2+·· 3 n-1=(3 (n+1)-3) 2 so an=(3 (n+1)-3) 2)+1