For the number series, An 1 5 2 2 An, how to use the pending coefficient to find the formula of the

Based on your example, I made an induction. Personally, I think it's the essence...... = =

In general, the formula that is transformed into an equal difference proportional series is a(n+1)=pan+r, where p,r are constants.

You can make it like this: a(n+1)+q=p(an+q) where q is to be determined, so that you make up a series of proportional proportions of equal difference, corresponding to your example (1).

Explanation: In this way, the sequence is a proportional series with "a1+q" as the first term and the common ratio is p, an+q=(a1+q)*p (n-1); n-1) is an exponent.

So an=(a1+q)*p (n-1)-q becomes an equal proportional series.

Explanation: a(n+1)=pan+r, the first term a1 in p, r, and an in this formula can be found by passing the specific question.

Next, the method of finding q: the formula a(n+1)+q=p(an+q) is transformed into a(n+1)=pan+pq-q, and the original formula a(n+1)=pan+r

Correspondingly: r=pq-q, then q=r (p-1), so q can be found according to p,r, which is a fixed value.

Therefore a(n+1)=pan+r, which can be turned into an equal difference proportional series, and the undetermined coefficient is used...

This is probably the general formula. Carefully understand the ...... you understand

Reason: The undetermined coefficient method has a general formula, and when you see this formula, you know that this method is used. This equation is a (n+1)=can+q

A simple analysis of the first field judgment and the song line, the details are shown in the Chong Kuan diagram.

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

a(n+1)+(n+1)-1 Divine Guess Lead Feast 2=-[a(n)+n-1 2].

a(n)+n-1 2} is the first proportional sequence of a(1)+1-1 2=1 2 and the common ratio is (-1).

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

Summary. The classmate should be an=n -n-56

The general formula for a known sequence (an) an=n -n=56(2) finds the value of n satisfying the inequality an<0.

The classmate should be an=n -n-56

Did you make a mistake?

2) an=n -n-56<0 i.e. -8

by an=an-1

2n gets an-an-1=2n

Then when n faction Cong Kuan Zheng Fan 2.

There is a2-a1=2x2

a3-a2=2x4

an-an-1=2n

Add the above n-1 equations.

an-a1=2(2+3+4+....)+n)

an-a1=2[(n-1)(n+2) dust bright2]an-a1=n2+n-2

So an=n2+n-1

When n 1, a1=1+1-1=1 also fits the above equation.

So the general formula.

is an=n2+n-1

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

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

1/[a(n+1) -1]=(2-an)/(an -1)=(1-an +1)/(an -1)=-1 +1/(an -1)

1 [a(n+1)-1]-1 (an -1)=-1, is a fixed value.

1/(a1-1)=1/(0-1)=-1

A series of numbers is a series of equal differences with -1 as the first term and -1 as the tolerance.

1/(an -1)=-1+(-1)(n-1)=-nan=-1/n +1=(n-1)/n

When n=1, a1=(1-1) 1=0, which is also satisfied.

The general formula for a series of numbers is an=(n-1) n.

Let's use the inductive method.

a1=1/2

a2=2/3

a3=3/4

So let an=n (n+1).

Obviously, both are true when n=1,2.

If n=n is true, then when n=n+1.

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

Q: The basic idea of the recursive (and initial or constrained) general terms of a known number series. A:

In the high school curriculum, the main emphasis is on the equal difference number series and the equal ratio number series; Complex problems are also solved by translating into both. It can be seen that the recursive formula of the equal difference series, the proportional series: an=a(n-1)+d; an=qa(n-1), both are first-order recursive relations (order:

i.e., the lower standard deviation of the unknown term in the formula), and its general form is an+xa(n-1)+y=0.It can also be converted to the following (*1).

Through a simple transformation, the solution of the recursive relationship of an+xa(n-1)+y=0 can be obtained, that is, the general term anExample: Known: xa(n) ya(n-1) z

Question: How do you construct a proportional sequence to find the general term a(n) and compare xa(n)-u=v(xa(n-1)-u) with xa(n) ya(n-1) z.

vx＝y,u-uv＝z

The solution: v y x, u z (1-v) xz (x-y) for functions where z is n, see the link given here. If it is a linear relationship between a(n+1), a(n), and a(n-1), it is called second-order linear recursion.

For second-order recursive, it can be solved by converting it to a first-order relation. This is exactly the same as when we study a quadratic equation and convert it into two linear equations. In view of this, people further summarize on this basis, and finally get rid of the transformation process, like the fixed formula of playing Go, summarize the method, get the formula, and then have the characteristic root method, and so on.