Number series questions, help, urgency!! The master is in!

a(2n) = f[a(2n-1)] = a(2n-1) +1,a(2) = a(1)+1=1+1=2.

a(2n+1) = g[a(2n)] = 2a(2n) +1,a(2n+2) = a(2n+1) +1 = 2a(2n) +1 + 1 = 2a(2n) +2,a(2n+2) +2 = 2[a(2n) +2],a(2n)+2} is the first proportional series with a(2)+2=2+2=4 and a common ratio of 2.

a(2n)+2=4*2^(n-1)=2^(n+1),a(2n) = 2^(n+1) -2, n = 1,2,..

a(2n+1) = 2a(2n) +1 = 2[2^(n+1) -2] +1 = 2^(n+2) -3,a(2n-1) = 2^(n+1) -3, n = 1,2,..

a(1)+a(2)+.a(2008)+a(2009)

a(1)+a(3)+.a(2009) +a(2)+a(4)+.a(2008)

n is an odd number. a（n+2）=g（a（n+1））=g(f(an)) =2（an+1）=2an+3

So a(n+2)+3= 2an-6=2(an+3), i.e., odd terms plus 3, is a proportional series with a common ratio of 2.

So a(2009)+3=2 1004*(a1+3)a(2009)=2 1006-3

Refer to the logarithmic function algorithm.

So there is ln2-ln1=ln2 1=ln2

The AN tolerance is d, and the BN total ratio is q

a2*b2=1 => (a1+d)*b1*q=1 => (a1*b1+d*b1)*q=1 => (1+d*b1)*q=1 => d*b1=1/q-1

a3*b3=(a1+2*d)*b1*q*q=(a1*b1+2*d*b1)*q*q=(1+2*d*b1)*q*q=(1+2*(1/q-1))*q*q=2*q-q*q

1- (q-1) squared.

I think it's minus infinity to 1

The front opens and the back closes.

It is to find a11, 9 times 2 to the 7th power.

a2+a5+a8=9,3a5=9

a5=3a3*a5*a7=-21

a5-2d)a5(a5+2d)=-21

a5)^2-4d^2]a5=-21

3^2-4d^2]*3=-21

3^2-4d^2=-7

4d^2=16

d^2=4d=±2

When d=2.

a5=a1+4d

3=a1+4*2

a1=-5an=a1+(n-1)d

5+2(n-1)

2n-71) when d = -2.

a5=a1+4d

3=a1+4*(-2)

a1=11an=a1+(n-1)d

11-2(n-1)

13-2n2) ( Sort out the recursive formula 2bn+1=bn+1 to get bn+1+1=2(bn+1), and then deduce that the sequence is proportional to 1 as the first term and 2 as the common ratio

According to the general term formula of (1) the number series }, and then find an, substitute the general term formula of cn= to obtain the number series, use the split term method to find the sum of the first n terms of the number series, and then find the range of n according to tn, and determine the minimum value of n Answer: solution: ( Prov:

From the meaning of the title, 2bn + 1 = bn + 1, bn + 1 + 1 = 2bn + 2 = 2 (bn + 1), and a1 = 2b1 + 1, b1 = 0, b1 + 1 = 1≠0, so the sequence is a proportional sequence with 1 as the first term and 2 as the common ratio ( Solution: From (1) knows, bn + 1 = 2n-1, an=2bn + 1 = 2n-1, so =

By , and n n*, the minimum n value that satisfies the condition is 10

(1)s(n)=(1/4)an^2+(1/2)an s(n+1)=(1/4)a(n+1)^2+(1/2)a(n+1)

Subtract the two equations to obtain a(n+1) = (1 4) a(n+1) 2-(1 4)an 2+(1 2)a(n+1)-(1 2)an

Simplified, a(n+1)-an=2, so an is the first equation of 2, and the tolerance is 2, and the difference series is an=2n

1. Use the point to substitute an and sn in the function as 1, and then write the formula of sn-1 as 2.

The relationship between an and an-1 is derived from the equation 1-2.

Use the relation of square difference to reduce all that can be reduced.

Finally, an-a(n-1)=2 is obtained

ps: Don't be intimidated by the question, it's actually very simple).

2. This is a typical dislocation subtraction method, because x is a real number, so it can be regarded as a known quantity (this method is used for multiplying the ratio by the difference).

Take a look at the process of pushing the textbook to the sum of the first n terms of the proportional series.

It's just that in the textbook, the constant series 1 (special difference series) is multiplied by the equal ratio of 3, and the method of scaling is used.

Pay attention to do it yourself, don't just want to see the answer, when you make it yourself, it will be very fulfilling, I wish you a successful solution to this problem!!

It's better to take a picture of the topic and send it up, you don't understand this symbol.

I studied it last semester, but I've almost forgotten it.

1 Total solution: 1) sn+s(n-1)=tan 2+2

sn+sn-an=tan^2+2

sn=tan^2/2+an/2+1

s(n-1)=ta(n-1)^2/2 + a(n-1)/2 +1

sn - s(n-1)=an=(t/2)[an^2-a(n-1)^2] +1/2)[an-a(n-1)]

t/2)[an^2-a(n-1)^2] -1/2)[an+a(n-1)]=0

1/2)[an+a(n-1)]*t(an-a(n-1))-1]=0

an>0 a(n-1)>0

t(an-a(n-1))-1=0

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

an=a2+(n-2)/t

a2=1/t

an=(n-1)/t

2) bn=1/[an*a(n+1)]=t²/[n(n-1)]=t²[1/(n-1)-1/n] n>=2

b1=ttn=t+t²[1-1/n]<2

Valid for any n n.

t²+t<2

20∴ 0

2a(n)a(n-1)+(n-1)a(n)=3na(n-1) divided by a(n)a(n-1) on both sides

2+ (n-1) a(n-1) = 3n a(n) Let b(n) = n a(n).

2+b(n-1) =3b(n)

3(b(n)-1) = (b(n-1)-1) So b(n)-1 is a proportional series, and the common ratio is 1 3

b(1) =1/a(1) = 2/3

b(n)-1 = (b(1)-1) (1 3) (n-1) gives b(n)= -(1 3) n+1

So a(n) = n (-(1 3) n+1).

Based on this, it is clear that the following is the ...... of (1-1 3)*(1-(1 3)2).1-(1 3) n) greater than 1 2 is sufficient.

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

Obviously, when n>=2, 2n+1<3n+3, and 1 [3 (2n+1)]-1 [1 3 (3n+3)] 0

So [1-1 3 (n+1)] 3 > (1-1 [3 (n)]) is 1-1 3 (n+1) >1-1 [3 (n)])1 3).

So there is 2 3 = (2 3) 1

.1-1/3^(n+1) >2/3)^(1/n)

Left (2 3) (1+1 3+1 9+1 27+...2/3)^(1/n)）

And 1+1 3+1 9+1 27+...2/3)^(1/n) = 3/2(1-(1/3)^n) <3/2

So left" (2 3) (3 2) = > 1 2

So the original formula is true.

a1=3/2,an=3na(n-1)/[2a(n-1)+n-1]

an=3na(n-1)/[2a(n-1)+n-1]

Take the countdown on both sides at the same time.

1/an=[2a(n-1)+n-1]/[3na(n-1)]

1/an=2/3n+(n-1)/[3na(n-1)]

n/an=2/3+(1/3)[(n-1)/a(n-1)]

n/an-1=1/3*(n-1)/[a(n-1)-1]

A sequence is the first proportional sequence with a common ratio of 1 to 3.

n/an-1=(1/a1-1)q^(n-1)

n/an-1=[1/(3/2)-1]*(1/3)^(n-1) (n>=1)

n/an-1=[2/3-1]*(1/3)^(n-1)

n/an-1=(-1/3)*(1/3)^(n-1)

n/an=1-(1/3)^n

an/n=1/[1-(1/3)^n]

an=n/[1-(1/3)^n]

A1A2....an<2n! .1)

i.e. (1 a1) (1 a2) ...1/an)>1/2n!

i.e. (1-1 3) 1*(1-1 3 2) 2*...1-1/3^n)/n>1/2n!

i.e. (1-1 3) (1-1 3 2) 2....1-1/3^n)>1/2 ..2)

First prove that when n n*, there is (1-1 3)(1-1 3 2)...1-1/3^n)>=1-(1/3+1/3^2+..1/3^n)..3)

The following is proved by mathematical induction.

Equation (3) holds when n=1.

This is true when it is assumed that n=k, ie.

1-1/3)(1-1/3^2)..1-1/3^k)>=1-(1/3+1/3^2+..1/3^k)

then n=k+1.

1-1/3)(1-1/3^2)..1-1/3^k) [1-1/3^(k+1)]>=1-(1/3+1/3^2+..1/3^k)[1-1/3^(k+1)]

1-(1/3+1/3^2+..1/3^k)-1/3^(k+1)+1/3^(k+1)(1/3+1/3^2+..1/3^k)

1-[1/3+1/3^2+..1 3 k+1 3 (k+1)] 3) is true.

Therefore, it is known by mathematical induction that Eq. (3) holds true for all n n*.

1-1/3)(1-1/3^2)..1-1/3^n)>=1-(1/3+1/3^2+..1/3^n)

1-(1/3)[1-(1/3)^n]/(1-1/3)

1-(1/2)[1-(1/3)^n]

1/2+1/2(1/3)^n

That is, equation (2) is established.

Thus equation (1) is established, i.e.

a1a2...an<2n!

See ** for details. I've made the first two questions, hehe.

(1) First of all, count down the two sides: 1 (an*n)=1 3*+2 3;

Then the undetermined coefficient method, let a number a such that there is n (an) + a = 1 3*, and we get -2 3 * (a) = 2 3, that is, a = -1;

Remembering bn=n (an)-1, then there is b1=-1 3 is not zero, so the sequence is a series of numbers with 1 3 as the common ratio, and the first term is -1 3.

bn=-1 3(1 3) (n-1)=-(1 3) n, so an=n*(3 n) [(3 n-1)].n=1,2,3...

2) First, there is 1-1 (3 n)=n an, and the two sides have [1-1 3]*[1-1 (3 2)]*1-1 3 n]=n!/(a1*a2*..

an), if true, there is [1-1 3]*[1-1 (3 2)]*1-1 3 n]>1 2, and mathematical induction proves: [1-1 3]*[1-1 (3 2)]*

1-1/3^n]>1-（1/3+1/3^2+..1 3 n) (this is the proof part, the proof is quite simple, so it is omitted) = 1-1 3 (1-1 3 n) (1-1 3) = 1 2 + 1 2 * (1 3 n) > 1 2, so the conclusion is valid!!