The number master comes to 10, and the number series master comes

Generally speaking, the general term of a series with only finite terms is infinitely many.

Because the general term n (n) and the function (n) are integer functions, if a general term is given, then there must be an integer function (n) of order of order -1 to satisfy the condition, so that as long as the order is greater than or equal to the number of general terms minus one, the integer function (n) may satisfy the condition, and only the corresponding undetermined coefficient is required. In fact, the Lagolange interpolation function is a special case!

Consider taking n= (n)=an*2+bn+c, then there is 9 27 a+b+c; 2/27=4a+2b+c;1/27=9a+3b+c;If the solution is a=4, then the fourth term is 9 27

If it is a third-order function n= (n)=an*3+bn*2+cn+d, it can also be obtained, but the expression in this case is not unique, it is infinitely many! So the answer to the fourth item is infinitely many, which is any number! Because as long as you write the fourth item, you can find the general term according to my method according to the first four items, and the topic will be satisfied!

You might as well take the fourth term as x, and x as any real number or even a complex number!

Let's take the third-order function n= (n)=an*3+bn*2+cn+d

Then we have 4 unknowns and 4 equations, and the expressions that can be found in general are functions with x as variables, and any x can be substituted to meet the problem!

If we increase the order of (n), then from linear algebra theory, we know that systems of equations must have solutions in general! So as to satisfy the topic!

One, observe, guess the result.

Second, imagine the sequence as a special function, solve the function with the undetermined coefficient method, and then find the unknown term.

Third, find the relationship between the front and back terms, and determine the general terms according to the relational formula.

Fourth, practice more, form feelings, and promote the use of the first method.

This question is very tragic, I guess the landlord was tricked or came to play us, and the third item of your question is wrong.

Let an=n 2 2*3 (n-1).

tn is the sum of an.

Apparently an is complicated: multiplying 2 by 3 becomes an an=n 2 3 n and then proving his sum tn is less than 3 2

When n>=42 n>=n 2, then write the first 3 terms, and zoom in from the 4th term.

.The sum of the proportional sequence is the use of the limit formula both.

And then add them up, apparently no, it's greater than 3 2

The conventional method doesn't work, it's weird.

This series can be summed directly, and it is found that the limit after summing is 9 4, so theoretically, even if you zoom in a little bit, you can't add a crack term by the method of pending coefficients.

Or multiply by 2 and divide by 3 (simpler) to become an=n 2 3 n and prove that his sum tn is less than 3 2

Above, but I couldn't find the data cable for my digital camera. Wait a minute.

The question written by LZ can only be understood by BT.

It is recommended that lz send the ** of the question.

It's better that way.

Might as well make a1=k

a2=k2+k

a3=k3+k2+k

a4=k4+k3+k2+k

a5=k5+k4+k3+k2+k

Observing the relationship between the two items of the necklace, we can see that an-an-1=kn where n>=2

where an is to the nth power of a, an-1 is the n-1 power of a, and kn is to the nth power of k).

The superposition method yields an=(an-an-1)+(an-1 -an-2)+a3-a2）+（a2-a1）+a1

Bring in an=kn+ kn-1 + k3+k2+k1=k(1-k n) (1-k).

Note that the general formula obtained above is under the condition of k>=2, and it is necessary to verify whether k=1 satisfies, that is, the result obtained by bringing k=1 into an=k(1-k n) (1-k) is also a1=k satisfied, so it can be summarized below.

an=k(1-k^n)/(1-k)

Each term of the sequence is the sum of a proportional series with the first term k and the common ratio k.

an=k(k^n-1)/(k-1)

This is the sum of a proportional sequence!

The difference between the first number -1 and the fourth number 3 is 2 = 4

The difference between the second number -1 and the fifth number 8 is 3 2 = 9

The difference between the third number 0 and the sixth number 25 is 5 2=25

It's all prime numbers. Every two numbers differ by the square of the prime number.

The difference between the fourth number 3 and the seventh number is 7 2 = 49

The seventh number is 52

After changing the title, it was much easier.

It is the nth number multiplied by 2 and the nth nth number is added

Is it a(s,t) or (t,s)?

Let's say a(s,t).

Then the first s-1 row has 1+2+4+8+......2 (s-2) = 2 (s-1)-1.

So this is the number 2 (s-1)-1+t.

So it is a=2[2 (s-1)-1+t]-1=2 s+2t-3

Isn't this the easiest number problem, you know that there are 2 (k-1) numbers in row k, then don't you know the expression for the first number in line k? If you don't know a(s,t) according to s, study hard, little brat.

(1) Because of the difference series.

Therefore, a5 a3=(a1+4d) (a1+2d)=5 9a1=-13d 2

s9/s5=(9a1+9*8/2d)/(5a1+5*4d/2)=11/(1+..n)=1/[(n+2)(n+1)/2]=2[1/(n+1)-1/(n+2)]

Therefore, the sum of the first ten terms is s=1+2(1 2-1 3)+2(1 3-1 4)+2(1/10-1/11)

A simple method: s9 s5=(2*9*(a1+a9)) 2*5*(a1+a5))=9*2*a5 5*2*a3=(a5 a3)*9 5=1

Question B: N term = 1 (1+2+3+..)n)=2/[n(n+1)]=2[1/n-1/(n+1)].

So the first n terms and.

sn=2(1/1-1/2)+2(1/2-1/3)+2(1/3-1/4)+.2[1/n-1/(n+1)]

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

2n/(n+1)。

So the sum of the first ten terms = 20 11

=(9/5)*(a1+a9)/(a1+a5)=(9/5)*(2*a5)/2*a3)

b.The question should be 1,1 (1+2).

n term = 1 (1+2+3+..)n)=2/[n(n+1)]=2[1/n-1/(n+1)].

So the first n terms and.

sn=2(1/1-1/2)+2(1/2-1/3)+2(1/3-1/4)+.2[1/n-1/(n+1)]

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

2n/(n+1)。

So the sum of the first ten terms = 20 11

Obviously, a(n)>0, so that b(n)=lg(a(n)), then.

b(n+1)=2b(n)+lg2.

b(n+1)+lg2]=2[b(n)+lg2]b(n)+lg2 is the proportional sequence, and then calculate b(n)+lg2, b(n), a(n)

1/a(n+1)=(2an+1)/3an=2/3+1/3an1/a(n+1)-1=(1/3)(1/an-1),1/a1-1=2/3

It is a proportional series with the first term 2 3 and the common ratio of 1 3.

1/an-1=(2/3)*(1/3)^(n-1)=2/3^n1/an=1+2/3^n=(2+3^n)/3^nan=3^n/(3^n+2)

There are many methods in the sequence, such as the accumulation method, a(n+1) an=1 n, the accumulation method, a(n+1)-an=n, the method of elimination of the split term, the method of 1 [n(n+1)]=1 n- 1 (n+1) dislocation subtraction, the sum of the proportional series, the method of the coefficient to be determined, and the general formula is generally obtained from the recursive formula.

This question cannot be asked so ... Method: This thing is basically based on need. The elimination method of item elimination, I see this in my memory only used in elementary school, from junior high school onwards into many specific methods.

Not sure what kind of answer you want