A high number of series, no, humbly ask for advice

an*a(n-1)=a(n-1)-an

Divide both sides by an*a(n-1) to get :

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

and bn=1 an

So, 1=bn-b(n-1) is a series of equal differences.

Then, bn=b1+(n-1)d=3+n-1=n+2tn=1 1*3+1 2*4+......1/n(n+2)=1/2[1-1/3+1/2-1/4+1/3-1/5……+1/n-1/(n+2)]=1/2[1+1/2-1/(n+1)-1/(n+2)]=3/4-1/2[1/(n+1)+1/(n+2)]

Because 1 (n+1)>1 (n+2), 1 2[1 (n+1)+1 (n+2)]>1 (n+2).

Then 3 4-1 2[1 (n+1)+1 (n+2)]< 3 4 - 1 (n+2).

So, i.e., tn<3 4 - 1 (n+2).

an*a(n-1)=a(n-1)-an

then an=a(n-1) (a(n-1)+1), so a1=1 3

a2=(1/3)/(4/3)=1/4

a3=(1/4)/(5/4)=1/5

And so on: an=1 (n+1).

So 1, bn=1 an=n+1

What does NBN stand for? Is it 1 (nbn) or (1 n) *bn wish you fun!

an=7sn-1+2

a(n+1)=7sn-1

Subtracting the two formulas yields a(n+1)-an=7an

Shift the term to get a(n+1)=8an

Therefore, an is a proportional series with 2 as the first term and 8 as the tolerance. an=2 3n-2, then bn=3n-2

tn=3n*n+(1+3n-2)*n/2

It's going to be a bad thing next... Friends and relatives leak.

1.a5*a5=a1*a17,a1+4d)*(a1+4d)=a1*(a1+16d),a1=2d, so an=2d+(n-1)d=(n+1)d

2.a1=2d, a5=6d, a17=18d, so the general formula for abn in the proportional series is abn=2d*3 (n-1).

Let bn=ai, i.e., 2d*3 (n-1)=(i+1)d, i=2*3 (n-1)-1, bn=2d*3 (n-1).

tn=c(1,n)b1+c(2,n)b2+c(3,n)+.c(n,n)bn

（2d/3）*∑c(j，n)*3^n*1^(n-j)]-2d/3 k=0,1,2,……n

It is obtained by the binomial theorem.

tn=(2/3d)(3+1)^n-2d/3

2d/3)*(4^n-1)

4^n+bn)/tn=[4^n+2d*3^(n-1)]/(2d/3)*(4^n-1)

Finding the limit gives lim(4 n+bn) tn=lim[2d 3+(3 4) n]=2d 3

So limtn (4 n+bn)=3 2d

You only judged b3>b2 based on the AN general term formula, and omitted b2>b1

Questions 1, 2, 3, and 5 are correct.

In question 4, q can take +2 and -2, and you missed -2.

Problem 6 is miscalculated, s3=2(1+q+q2)=26, solve q=3 or -4, and the rest of an can be brought into the solution by yourself.

Question 7 because an is a proportional sequence, so a1a3=(a2) 2, bring in a1a2a3=8 to get a2=2, and we know a1+a2+a3=-3, so 2 q+2+2q=-3, the solution is q=-1 2 or -2, so a4 will naturally be found, and I will not count it.

Question 8: a5a9=(a7) 2, bring in the data to get a9=9

Question 9 a5-a1=a1(q 4-1)=15,a4-a2=a1(q 3-q)=6, divide the two formulas, then eliminate a1, get (q 2+1) q=5 2, so you can solve q, and then bring in any of the above formulas, you can find a1, natural a3=a1*(q 2), ask for it yourself, I don't ask for it.

Question 10 I can't understand your answer, my answer is q 5 = a9 a4 = 243, I get q = 3, so an=4*[3 (n-4)].

Question 11 is the same as question 10, first seek the common ratio Q, calculate it yourself, I won't count it anymore, if you really won't ask me again.

This topic is so confusing, how can I wait for the difference, and I can't figure out which one is the difference and which is the same.

Since Question 12 gives linear recursion, why is it still said to be a series of equal differences? If you only use the eigenequation to solve the condition of the recursive relation, you shouldn't be in high school yet, right? The eigenequation is a method in the math competition, which can solve all linear recursive equations, and you can study it if you are interested, but I won't talk about it here, it's too tiring.

Question 13 is obviously positive for the first 12 items, 0 for the 13th item, and the rest are all negative, and the biggest one is s13 with common sense.

Hungry, what kind of number sequence is question 14, I can't think of how to do it with such conditions.

If I don't write it, I'm almost exhausted, these questions are very basic number series problems, and I guess you haven't learned these things yet, so it may be difficult to do; I am a graduate of the third year of high school this year, anyway, the advice to you is to memorize more than 20 formulas related to the sum and equal difference, equal ratio and sum, and then master the basic methods, as well as the method of finding the general term of the characteristic root, then it is more than enough to cope with the college entrance examination.

If the above topic is still really unclear, my QQ is 873650215

s is the limit of the sn.

It's easy to find this with a formula.

s=a1/(1-q）=3/2

To ks sn constant is established, due to the increase of sn increments.

So as long as ks=s1, i.e. the maximum value of k is 2 3

S4-A1=28 gives A2+A3+A4=28, and A3=A2Q, A4=A2Q 2, and A2=4, so vertical pants 4+4Q+4Q 2=28

q=2 or -3(an>0, so it is not compatible with the rest of the house) a(n-3) an=1 q 3=1 8

It's too simple, fast, and hidden.

This is done by dislocation subtraction.

tn=b1+b2+..bn

1/2*tn=。。Huai Sun.

It just sorts it into a series of numbers with the same mother and a numerator difference of 2.

Subtract to get a proportional series + an extra number.

The solution is over.