A question about finding a general term for a series of equal differences question below .

It is known that f(x)=a x+a x +a x +a n x , and a , a , a , a , .,a n is a series of equal differences, n is a positive and even number, and f(1)=n, f(-1)=n; Finding the general term of a n?

Solution: f(1)=a +a +a +a+a n = n (n is an even number).1)

i.e. na + (1 2)n(n-1)d = n ; Therefore a = n-(1 2)(n-1)d;

f(-1)=-a₁+a₂-a₃+.a‹n›=n...2)

1) + (2) to get 2 (a + a + a +a 2k )=n +n, where 2k=n, k=n 2;

That is, there is 2[ka + (1 2)k(k-1)(2d)]=2ka +2k(k-1)d=2(n 2)a +2(n 2)(n 2-1)d=na +n(n 2-1)d=n +n

a₂+(n/2-1)d=n+1；Therefore, a = n+1-(n 2-1)d;

a -a = [n+1-(n 2-1)d]-[n-(1 2)(n-1)d]=1+(1 2)d=d, so d=2;a₁=n-(n-1)=1;

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

Note: Your approach is: since s n = n; Therefore s n-1 = (n-1) ; So a n = s n -s n-1 =n -(n-1) =2n-1

This should be possible. Since it is a series of equal differences, and n is an even number, then the sum of the first n terms is n; n-1 is an odd number, then the sum of the first (n-1) terms is (n-1) ; Because n is even, n 2, and when n 2 a n = s n -s n-1 = n - (n-1).

2n-1；Then check a = 1 when n = 1 and a -a = 3-1 = 2 = d, so a n = 2n-1 is also correct when n = 1.

There's nothing wrong with that, you teachers may be a little confused.

The disadvantage of this method is that it does not make full use of the conditions given by the question, and it does not quite meet the requirements of the questioner.

When x=1, we can know f(1)=sn, but there is no way to find how much s(n-1) is equal to according to the meaning of the question.

In this problem, we must first find the tolerance according to f(-1)=n.

a2-a1）+(a4-a3)+.an-a(an-1)]=nd*n/2=n d=2

Then find the first term according to f(1)=sn=n2 =na1+n(n-1)d 2.

Finally, an is found according to the first term and tolerance

First of all, my intuition is to substitute, and then observe a1+a2+a3...=n2 -a1+a2-a3+a4..=n According to this clause, it can be concluded that the post-ante-ante=a constant is 2, so the tolerance comes out.

sn=na1+n(n-1)d/2

sn=na1+n2-n=n2 so a1=1,, is the use of the sn method you said to treat anxn as a general term of a series? In this way, because xn can't be approximated, an

The tolerance is 3 for the blind calendar

So an -2+3(n-1) 3n-5

The tenth item is A10 3*10-5 Lao Sou 25

The sum of the first ten terms is (-2+25) Shen Min*10 2 115 I hope the explanation is clear enough

The tolerance is 3

So an -2+3(n-1) 3n-5

The tenth item is A10 3*10-5 25

The sum of the first ten terms is (-2+25)*10 2 115, I hope the explanation is clear enough

Known sequences. is an equal difference in the number of wax carrying columns, and.

1) Find the number series.

the general term formula; 2) Verification:

2) Reference analysis.

Test question analysis: (1) Because of the number series.

is a series of equal differences, and.

By listing the corresponding equations under these conditions, the first term and tolerance of the equal difference series can be found, and the series can be found.

, you can find the series.

The key to this problem is the understanding of a more complex number sequence, and the logarithmic operation is also prone to error.

Because from (1) to the series.

The general formula of the term is to find the number series according to the needs of the topic.

The first n terms and formulas, so the general term formula can be found by the general repentance calculation, and then the conclusion can be obtained by using the summation formula of the proportional series.

Analysis of test questions: (1) Let the tolerance of the difference series be d, which is corrected.

Get. So d=1;So. Namely.

2) Proof of: So.

1)an=sn-sn-1

Bring in the original formula, sn-sn-1+2sn*sn-1=0sn-1 -sn=2sn*sn-1

So (1 sn)-(1 sn-1)=2

So it's proportional.

Because 1 s1 = 1 a1 = 2 so = 2n

an=sn-sn-1

1/2n）- 1/2（n-1）

1/[2n（n-1）]

an=sn-sn-1

an+2sn*sn-1=0

sn-sn-1+2sn*sn-1=0

Divide by sn*sn-1

1/sn-1/sn-1=2

1 sn) is a series of equal differences.

s1=a1=1/2

1/sn=2n

sn=1/2n

an=sn-sn-1=-1 (n-1)n,n2n=1,a1=1 2

The tolerance is 3

So an -2+3(n-1) 3n-5

The tenth item is A10 3*10-5 25

The sum of the first ten terms is (-2+25)*10 2 115, I hope the explanation is clear enough

The sum of any series of equal differences can be expressed as.

1/2(a1

an)*n, where a1 is the first number, an is the nth number, and n represents the number of the number of the series, then the sum of the first 4 terms of the equal difference series is 2

There are 1 2 (a1

a4)*4=2

a1a4=1---1)

The sum of the first 9 terms of the difference series is -6

There are 1 2 (a1

a9)*9=-6

3a13a9=-4---2)

From (1) and (2), we can get a9-a4=-7 3, then we can find the tolerance of the difference series as d=-7 3 (9-4)=-7 15, then a4=a1-7 15*3=a1-7 5

Substituting (1) yields a1=6 5

Then the nth term of the series is.

an=a1n-1)d=6 5-(n-1)7 15=5 3-7n 15 then has its first n terms and for.

1/2(a1an)*n

5/3-7n/15)*n

n(43-7n)/30