Senior 2 math series questions, high school sophomore math series questions

For the first question, choose A

Not all sequences have a common term.

For example: Prime number (prime number) column: 2, 3, 5, 7, 11, 13, 17 ,..For thousands of years, many Chinese and foreign scholars have tried to find its general formula, but they have not found it.

In addition, it is impossible to have a common term if you arrange an infinite number of numbers without any regularity in a certain order.

Question 2: The sum of the first 2 terms equals the third term, so x=21 Question 3: is an infinite sequence of numbers, and all terms are equal to 4

Question 4: The denominator of an is rationalized and obtained by subtracting n under the root number n from the root number, so a1+a2+a3+a4+.an=2 under the root number - 1 under the root number + 3 under the root number - 2 under the root number + 4 under the root number - 3+ under the root number.

n+1 under the root number - n under the root number, so a1+a2+a3+a4+.an=n+1 minus 1 from the root number, so n=100

The first question will not be I get an A

Question 21

Question 3: I guess so.

Question 4: The denominator of an is rationalized and obtained by subtracting n under the root number n from the root number, so a1+a2+a3+a4+.an=2 under the root number - 1 under the root number + 3 under the root number - 2 under the root number + 4 under the root number - 3+ under the root number. n+1 under the root number - n under the root number, so a1+a2+a3+a4+.

an=n+1 minus 1 from the root number, so n=100

It is known to be an equal proportional series, a2 = 2, a5 = 1 4

Let the common ratio be q, then a5=a2*q 3

So q 3 = a5 a2 = 1 8 so q = 1 2 can be deduced from being a proportional series.

The common ratio is an*a(n+1) (a(n-1)*an)=a(n+1) a(n-1)=q 2=1 4

The first term is a1*a2=a2*a2 q=8

So a1*a2+a2*a3+.an*a(n+1)=8*(1-(1 4) n) (1-1 4)=32*(1-(1 4) n) 3 (using the equation for summation of proportional sequences).

(an+1)-1=1 [2-an]-1=(an-1) (2-an), let bn=an-1, then b(n+1)=bn (1-bn), 1 (bn+1)=1 bn-1,1 (bn+1)-1 bn=-1, so the first term is 1 b1=1 (a-1), the tolerance is -1 equal difference series, 1 bn=1 (a-1)-(n-1), bn=(a-1) [n(1-a)+a], so an=bn+1=[( n-1)(1-a)] [n(1-a)+a]

You should re-enter the question, in this case, you can't calculate the problem when you do the math. Is it n+1 at the bottom of A or n plus 1 at the bottom of A?

1. Calculate a few: a2, a3, a4,。。

2、an=[(n-1)-(n-2)a1]/[n-(n-1)a1]（n>=2）

3. The inductive method can be proved.

It is easy to know from the recursive formula, an>0 (square to positive);

Substituting a1 and a2, the test and calculation are carried out, and it is found that an is decreasing and not equal to 0;

Combined with convergence, it is known that the series converges to 0 when it tends to positive infinity, so there are infinite terms in the series.

a(n+1)=an (1+3an) (n+1) means the subscript 1 a(n+1)=(1+3an) an (reciprocal on both sides of the equation)1 a(n+1)-1 an=3

Therefore, it is a series of equal differences with 1 a1 = 1 as the tolerance of prime minister d = 3 1 an=1+3(n-1)=3n-2

So an=1 (3n-2).

It's a holistic idea!! I don't understand hi me.

Solution: Take the reciprocal: 1 an+1=(1+3an) an=1 an +3

So: the series is a series of equal differences with a tolerance of 1 digit and a tolerance of 3 digits, so: 1 an=3n-2, so, an=1 3n-2

Because a(n+1)=an 1+3an

So 1 a(n+1)=3+1 an

So 1 a(n+1)-1 an=3, and because a1=1, 1 a1=1, a2=1 1+3=1 4, 1 a2=4

Therefore, 1 an is a series of equal differences with 1 as the first term and 3 as the tolerance, so 1 an=1+(n-1)*3=3n-2, so an=1 (3n-2).

You can count down both sides and you can find the pattern.

There are 2n+1 terms, and the common ratio is x, so s=[1-x (2n+1)] 1-x) because it is an equal proportional series, so a3*a9=a5*a7=32, and a3+a9=18, so the solution is a3=16,a9=2 or clear fiber block a3=2,a9=16 because it is known that the proportional vertical column is an increasing series, a1>0, q>1, so it can only be a9=16, a3=2

Q 6=a9 a3=8,q= 2,a1=1sn=[1-2 (n 2)] 1- 2)=(2+1)*[2 (n 2)-1].

...... by a1+3a2+3 2a3+3 (n-1)an=n3 and a1+3a2+3 2a3+......3 (n-1)an+3 na (n+1)=(n+1) 3.

3^n*a_(n+1)=1/3

So a (n+1)=1 [3 (n+1)] so an=1 (3 n)=

So bn=n*3 n

Let its first n terms and be s

then s=3+2*3 2+......n*3^n

3s=3^2+2*3^3+……n-1)*3^n+n*3^(n+1)

The above two equations are subtracted from each other.

1-3)s=3+3^2+3^3+……3^n-3^(n+1)=[3^(n+1)-3]/2+3^n-3^(n+1)=3^n-[3^(n+1)+3]/2

So s=[3 (n+1)+3]-2*3 n

In this question [ ], the tolerance is set for the lower foot mark, and the tolerance is d

by a1+2a2+3a3+.nan=bn*n(n+1) 2 (1).

Get a1+2a2+3a3+.n-1)a[n-1]=b[n-1]*(n-1)n 2 (2).

a1+2a2+3a3+..nan+(n+1)a[n+1]=b[n+1]*(n+1)(n+2) 2 (3).

Formula 1 - Formula 2 gives nan= 2 i.e. an=(nd+bn+b[n-1]) 2 (4).

Formula 3 - Formula 1 gives (n+1)a[n+1]= 2, i.e. a[n+1]=(nd+2b[n+1]) 2 (5).

From Equation 5 - Equation 4 yields a[n+1]-an=(nd+2b[n+1]) 2-(nd+bn+b[n-1]) 2

2b[n+1]-bn-b[n-1])/2

3d 2 is constant.

So it's a series of equal differences.