Two questions about the number series. Two number series questions

First. Solution: Because a(n+1)=a(n)+1 (n(n+1)), so a(n+1)-a(n)=1 (n(n+1)) so a(n)-a(n-1)=1 (n(n-1))a(n-1)-a(n-2)=1 ((n-1)(n-2))a(2)-a(1)=1 (2*1).

Add all of the above to get it.

a(n)-a(1)=1/(2*1)+1/(3*2)+…1/(（n-1）(n-2))+1/(n(n-1))+1/(n(n+1))

i.e. a(n)-1=1-1 2+1 2-1 3+1 3-1 4+......1/（n-1）-1/（n-2）+1/n-1/（n-1）

i.e. a(n)-1=1-1 n

So a(n)=2-1 n

So a(10) = 19 10

I didn't understand the second question......

Based on known conditions.

a2-a1=1/1*2

a3-a2=1/2*3

a4-a3=1/3*4

a5-a4=1/4*5

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

So superimposed. a(n+1)-a1=1/1*2+1/2*3+1/3*4+..1/n(n+1)

The items in the back are regular, oh, take them apart.

a(n+1)-a1=1-1/2+1/2-1/3+1/3-1/4+..1/n-1/n+1

Anything that can be eliminated in the middle is eliminated.

Finally, a(n+1)-a1=1-1 n+1

So a10 = 2-1 10 = 19 10

I can't understand the back -

Question 1. a(n+1)=a(n)+1/n-1/(n+1)a(n+1)+1/(n+1)=a(n)+1/n=...=a1+1=2a(n)=2-1/n

a(10)=2-1/10=19/10

Question 2. It should be a(n)=(2n-1)+2 n, otherwise it makes no sense to write a constant 3 as -1+2 2

sn=(sum of a series of equal differences) + (sum of a series of equal differences) n 2+2 (n+1)-2

1、a1=1

a2=a1+1/1*2=1+1/1*2

a3=1+1/1*2+1/2*3

an=1+1/1*2+1/2*3+……1/(n-1)n1+(1-1/2)+(1/2-1/3)+…1/(n-1)-1/n)1+1-1/2+1/2-1/3+……1/(n-1)-1/n2-1/na10=2-1/10=

2. There must be something wrong with the question, otherwise it will be much simpler.

1)f'(x)=2x+3

a(n+1)=2an+3

So laugh at a(n+1)+3=2(an+3).

an+3} is the first term a1+3=3+a, and the common ratio is 2.

an+3=(3+a)2^(n-1)

2) an=s(n)-s(n+1) can be obtained

and s(n)=1+3+5+....+2n-1)=n 2, so an=n 2-(n+1) 2=-2n-1

Question 1: The numerator and denominator on the right are divided by an, a(n+1)=1 and swiftly open(1 an+n 2).Then take the reciprocal at the left and right ends and then shift the phase, you can get 1 [a(n+1)-1 an]=1 n 2, and then you can add the left and right ends at the same time, and finally the mu is left with 1 a(n+1)-1 a1=1+2 2+...

n 2, the second question: just turn the virtual n into n-1, and then divide it!!

The first series feels like something is wrong, so let's check it again!

The second brings n=n+1 into a1*a2*a3....an*a(n+1)=(n+1) 2, and dividing the mu by the disturbance beats on both sides of the conditional equation yields a(n+1)=(1+1 n) 2, then an=(1+1 n-1) 2(n>1).

S40 Explanation: Let the common ratio of the original series be q, note.

S10 is a, and the 10th power of Q is q, then there is.

s30=a+a*q+a*q^2

Thereupon. s30-a)/a=q+q^2=6

To solve the one-dimensional two-dimensional equation on the right, and since q is equal to the 10th power of q and is greater than zero, the negative root is taken as q=2

Thereupon. s40 = s30 + a * q 3 = 70 + 10 * 2 3 = 150 in its call to the celery, the symbol a b means a to the b power).

Because s1 = 10 and s30 = 70, s20 = 60

And because s10:s20=s30:s40

So s40=240

Hello! s40

Let the common ratio of the original series be r, denote s10 as a, and denote r to the 10th power as r, then there is.

s30a+a*r+a*r*r

So (s30-a) ar+r*r

To solve the one-dimensional two-dimensional equation of the right-click side of the panicle book, and because r is equal to the 10th power of the guess bright macro r is greater than zero, the positive root is taken.

r=2 next, s40

s30a*r*r*r=70

If it helps you, hope.

The double sequence 11 7 5 3 is a prime guess sequence, and 18 12 6 is a multiple of 6.

1. I made a few mistakes, the number series is 1 4, 3 4, 5 4, 7 4, and the tolerance is 1 2

Using Veda's theorem, x1+x4=2, x2+x3=2, and x1=1 4 are obtained.

2. (1) The construction method constitutes an equal proportional sequence, which is divided into two steps, one to deal with n and one to deal with constants.

a(n+1)+2(n+1)=2a(n)+4n-1

a(n+1)+2(n+1)-1=2[a(n)+2n-1]

Therefore, the sequence b(n)=a(n)+2n-1 is an equal proportional series, the first term b(1)=3, and the common ratio q=2

Therefore, b(n)=3*2 (n-1), a(n)=3*2 (n-1)-2n+1

3、d(n)=ln[c(1)c(2)……c(n)] n is a series of equal differences.

4、b(1)b(2)……b(n)=b(1)b(2)……b(17-n) （n<17）

Consider b(8)b(10)=b(7)b(11)=....b(1)b(17)=1

The second question and the second question have not been made yet, and then think about it, send these few first, these 100 points are really hard to earn, hehe.

Let me add the idea of 2 questions, 2 questions, and 3 and 4 questions.

2. (2) The right side of the equal sign is the fractional form of a(n), so considering the construction of a reciprocal proportional series, it is also divided into two steps, one step to find the constant on the denominator, and the second step to find the common ratio and constant terms.

Let x make a(n+1)+x=[5a(n)+4] [2a(n)+7]+x=[5a(n)+4+2xa(n)+7x] [2a(n)+7].

There is an equation x(5+2x)=4+7x, and the two roots -1,2 are solved, and the answer on the first floor is to take -1, and to do it down is to bring in x and then take the reciprocal.

3. Because the exponent of the common ratio in the proportional series is equal difference, it is necessary to find the equal difference series from the proportional series, and you can consider taking the logarithm.

4. b(9)=1, so the product of the first and second terms corresponding to the 9th term is 1; The difference series is the sum of the corresponding terms before and after is 0. Generally, when encountering equal differences, the sum is considered, and the equal ratio is considered the product.

1.Let the four roots be a, b, c, d, a2(1) From the recursive relation, a(n+1)+2(n+1)-1=2(a(n)+2n-1), so that b(n)=a(n)+2n-1

Then b(n+1)=2b(n), it is not difficult below.

2) 1 (a(n+1)-1)+1 3=3(1 (a(n)-1)+1 3), order from the recursive relation.

b(n)=1 (a(n)-1)+1 3, then b(n+1)=3b(n), and then sit down.

The secondary root number (c(1)*c(2)*....c(n)) is a proportional series.

n<17

d=7/4，c=3/4，d=5/4

a(n+1)+2(n+1)-1b(n)=1 (a(n)-1)+1 3, then b(n+1)=3 b(n1 (a(n+1)-1)+1 3=3(1

d(n)=n root number(c(1)*c(2)*....c(n)) is a proportional series.

a(n)-1)+1/3),=2（a(n)+2n-1）.b(1)*b(2)*…b(n)=b(1)*b(2)*…b(17-n),n<17

Hello, I am helping you to inquire about the relevant information and will reply to you immediately!!

Here we have found the relevant answer for you: the general formula for the difference series is a n=a 1+(n-1)d, so a (i+2)-a i=a 1+(i+2-1)d-[a 1+(i-1)d]=2d is true in all cases. And then the other way around, a (i+2)-a i=2d does not guarantee a series of equal differences, e.g. 0, 0, 2d, 2d, 4d, 4d, etcThis is true, but it is clearly not a series of equal differences.

an+an+1-[a(n-1)+an]=d+d=2d(n is greater than or equal to 2), do you understand? Specializes in high school math.

Basic Measurement Method.

Because of the proportional series, the formula an=a1*q n-1a5 a4=a1q 4 a1q 3 =q is used

Similarly, a3 a2=q

Again, a2 a1=q

Because q is a constant (considered known).

So a1 a2=a3 a2=a4 a3=a5 a4 so a1, a2, a3, a4, a5 are in equal proportions.