When proving the difference and the ratio series, when should a1 be taken out separately?

In order to express the rigor of your thinking, it should be.

an-an-1 =2, which should be the condition of n>=2, but it doesn't seem to have anything to do with what you said about a1.

Because n=2 is a1, it is logically feasible.

Your example doesn't illustrate the problem you asked, and only when n can't get 1 satisfies the relation, it needs to be alone.

Consider A1

No, you don't. Although n-1》1,n》2 here, when n takes 2 is a2-a1=2, it is obvious that n cannot take 1, because there can be no a1-a0.

To put it bluntly, you can only take n in this formula"2

an - an-1 =2

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

a2-a1=2

The derivation process is to add the left side and the right side to the right at the same time.

an-a1=2（n-1）（n>=2）

Because this takes out the a1.

Knowing a1, find an

Then substitute 1 into an to see if it is the same as the known a1, if it is the same, an does not change, if it is different, it is written separately.

When the equation given is sn=f(an).

It needs to be classified, a1=s1, n>=2, an=sn-s(n-1) when the value of a1 is given, and the recursive an=f(a(n-1)), n>=2 is given.

There is no need to classify ......

The formula for finding the difference series a1 is: {an}=a1+(n-1)d. A series of equal differences refers to a series of numbers in which the difference between each term and its previous term is equal to the same constant from the second term, which is often denoted by a and p.

This constant is called the tolerance of the equal difference series, and the tolerance is often denoted by the letter d.

A sequence of numbers (sequenceofnumber) is a function that defines a domain with a set of positive integers (or a finite subset of it). Each number in the sequence is called an item in the sequence. The number in the first place is called the first term of the series (usually also called the first term), the number in the second place is called the second term of the series, and so on, the number in the nth position is called the nth term of the series, which is usually denoted by an.

The formula for finding the difference series a1 is: {an}=a1+(n-1)d. A series of equal differences refers to a series of numbers from the second term onwards, in which the difference between each term and its previous term is equal to the same constant, which is often expressed by a and scattered p.

This constant is called the tolerance of the equal difference series, and the tolerance is often denoted by the letter d.

A sequence of numbers is a function that defines a domain with a set of positive integers (or a finite subset of it) as a function, and is a sequence of ordered numbers. Each number in the sequence is called an item in the sequence. The number in the first place is called the first term of the series (usually also called the first term), the number in the second place is called the second term of the series, and so on, the number in the nth position is called the nth term of the series, which is usually denoted by an.

The sequences A1 and A2 have the middle term of the land shout.

The equivalence of a1 and a2n-1 is 1.

Solution: is an equal difference series, the equal difference middle term between A1 and early vertical field A2 is 1, and the equal difference middle term between A2 and A3 is 2.

a1 + a2 = 2, a2 + a3 = 4, subtract the two formulas to obtain a3-a1 = 2d = 4-2.

The solution is d=1.

Yes, the equivalence of a1 and a2n-1 is 1

No, the middle term is the middle term of the three terms that are equally distanced before it becomes the middle term.

The general formula for the random kernel column of the hand difference in the same book is: an=a1+(n-1)d

The first n terms and formulas are: sn=na1+n(n-1)d2 or sn=n(a1+an)2

The above formula can be obtained by anti-state suspicion.

Because an=sn-s(n-1) can only be true when n 2 is true, this formula can be used to solve the problem.

When n=1 s0 doesn't make sense.

If you test a1, you only need to use an=sn-s(n-1) to solve an=1 and then get n=1

See if A1 will be equal to A1 as told in the title

If it is, then it can be concluded that an=....

If not, then write a1=...an=...n≥2）

In fact, there is no case where a1 does not conform to an in the equal difference series.

It is not necessary to test A1

an=sn+1 -sn？？

That's not right...

an+1=sn+1 -sn

In fact, it is still necessary to test A1.

Although an+1=sn+1 -sn is true for a1.

But the solution is that an+1 n belongs to n+

Then this formula can only start from n=2, so you still have to test a1, and this formula is actually the above formula of the exchange of yuan, there is no difference in essence, I hope it will help you.

Sometimes A1 doesn't fit the path formula

1. A11 A10<-1<0, so A11 and A10 are on both sides of the origin.

2、|a11|/|a10|>1, so |a11|>|a10|, i.e., a11 is farther away from the origin than a10.

3. Because A11 and A10 both may be positive and negative, A11 may also be larger than A10 or smaller than A10, so the tolerance may be greater than 0 or less than 0

4. The absolute value of the tolerance is equal to |a11|+|a10|