Who has a number series problem composition tips classic question oh 20

The sum of the first n terms of the equal difference series is sn

a3=10, a2,a4,a7 in proportional series.

Establish. an =a1+(n-1)d

a3=10a1+2d=10 (1)

a2, a4, a7 in proportional series.

Derivation. (a4)^2

a1+d)(a1+6d)=(a1+3d)^2(a1+2d-d)(a1+2d+4d)=(a1+2d+d)^2(10-d)(10+4d)=(10+d)^2100+30d-4d^2=100+20d+d^25d^2-10d=0

d^2-2d=0

d = 2 by equation (1).

a1+2d=10

a1+4=10

a1 = 6.

an=a1+(n-1)d = 6+2(n-1) =2n+4sn=a1+a2+..an

n(a1+an)/2

n(6+2n+4)/2

n(n+5)

Write bn = 2 (sn+6) and find the first n terms of the sequence and tnbn = 2 (sn+6).

2/(n^2+5n+6)

2/[(n+2)(n+3)]

2[ 1/(n+2) -1/(n+3)]tn=b1+b2+..bn

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

(1) Let the first item be a and the tolerance be d, according to the conditions.

a₁＋2d＝10，①

a 3d) a d)(a 6d), solution a 6,d 2,(round off a 10,d 0) so an 2n 4,sn na n(n 1)d 2 n 5n.

2) bn＝2/(n²＋5n＋6)

2/[(n＋2)(n＋3)]

2[1 (n 2) 1 (n 3)], so tn 2[1 3 1 4 1 4 1 5 ...1/(n＋2)－1/(n＋3)]

2[1/3－1/(n＋3)]

2n / 3(n＋3)]

Solution: (1) a2, a4, a7 into an equal proportional series, then there is a4 a2*a7; Since a2, a4, and a7 are elements in a series of equal differences, then a2 a1 d, a4 a1 3d, a7 a1 6d;

Synoptic, available, a1 9d 6a1*d a1 7a1*d 6d ; and because a3 10 a1 2d; ③

Then there is 3d a1*d 0;

3d＝a1，④

United Links d 2,a1 6;

sn＝na1＋n（n－1）*d/2＝5n＋n²；

bn＝2/（n²＋5n＋6）＝2/［（n＋2）（n＋3）］＝1/（n＋2）－1/（n＋3）

The final result is then calculated by accumulation.

tn＝n/（3（n＋3））

Personally, I think to do a number series problem:

Step 1: Master all the formulas related to the number series (general formulas, summation formulas, etc.), and the commonly used properties.

Step 2: Do the questions to master the rules and summarize the question types. Common question types such as finding a formula for general terms, summing formulas, proving that a series of numbers is a series of equal differences or proportional numbers, proving that an inequality is true, and so on. Summarize it in a small notebook.

Step 3: Problem training, that is, use the template in your mind (the question type summarized before) to do the problem.

Step 4: Experience the experience and expand and extend. At this point, just count the modules, and you've reached the realm of perfection.

In my experience, there is a one-size-fits-all method for number series problems, and that is the "basic measurement method".

I wish you success!

1) From the meaning of the title, it can be known.

a1=4，an=a1*q^(n-1)=4*q^(n-1)，sn=a1*（1-q^n）/(1-q)=4*1-q^n）/(1-q) (q≠1,0)

2s2=s3+s4；2*a1*(1-q 2) (1-q)=a1*(1-q 3) (1-q)+a1*(1-q 4) (1-q).

Cancel the common divisor to get 2*(1-q 2)=(1-q 3)+(1-q 4).

2*q^2=q^3+q^4

2 =q+q^2

Because q ≠ 1,0, q = 2

an=4*（－2）^(n-1)

2) From the meaning of the title, it can be known.

bn=log(2)|an|=log(2)|4*（－2）^(n-1)|=log（2)（4*2^（n-1）)=log(2)(2^(n+1))=n+1

tn=1/b(1)·b(1+1)+1/b(2)·b(2+1)+…1/b(n)·b(n+1)

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

tn=1/2*3+1/3*4+……1/(n+1)*(n+2)

tn=(1/2-1/3)+(1/3-1/4)+…1/（n+1）-1/（n+2））

tn=1 2-1 (eggplant chain n-2).

s3, s2, s4 into equal difference friends = >

a1(1+q) *2= a1(1+q+q^2)+a1(1+q+q^2+q^3)

q=-2 or q=0 (rounded) to provoke.

an=a1*q^(n-1)=4*(-2)^(n-1)=(2)^(n+1)

bn=log(2)|an|=log(2) |2|(n+1)=n+1tn= s =s=1 b1 - 1 b(n+1)=1 2 - 1 lead only (n+2).

1) Lisen Liang a1 = 4 a2 = -8 a3 = 16 an=(-2) n + 1 limb Yun Fang Chunna.

2）bn=n+1 tn=1/2-1/(n+2)

Hello landlord. First question:

Let the tolerance be d and the tolerance be q

then b2=q,s2=a1+a2=3+3+d=6+d,b3=q 2,s3=a1+a2+a3=3+3+d+3+2d=9+3d

B2S2=Q(6+D)=64, B3S3=Q2 (9+3D)=960, i.e. Q2 (3+D)=320

From q(6+d)=64, we get 6+d=64 q, then 3+d=(64 q)-3, so q 2 (3+d)=q 2 (64 q)-3q 2=64q-3q 2=320, so q=40 3 or q=8, but when q=40 3, d is negative, so it is rounded.

So q = 8, d = 64 8-6 = 2

So an=3+2(n-1)=2n+1, bn=1 8 (n-1)=8 (n-1).

Second question: sn=(3+2n+1) n 2=n (n+2).

So 1 sn=1 [n (n+2)]=1 2) (1 n-1 (n+2)).

So 1 s1 + ......1/sn=1/2×(1/1-1/3+1/3-1/5+……1/n-1/(n+2))=1/2)×(1-1/(n+2))=n+1)/(2n+4)

Hope you're satisfied.

Let the tolerance be d, the tolerance is (6 d)*q=64, b3s3=960=>(9 3d)*q 2=960=>(3 d)*q 2=64q-3*q 2=320, q=40 3, d=rounded); q=8,d=2。So an=2n 1 and bn=8 (n-1). 1/s1 1/s2 ……1/sn=1/3 1/((3 5)*2/2) 1/((3 7)*3)/2 … 1/((3 2n 1)*n)/2=1/3 1/((3 5)*2/2) 1/((3 7)*3)/2 … 1/((2 n)*n)=（1/2)*(1-1/3 1/3-1/5 1/5-1/7 … 1/n-1/(n 2))=1/2)*(1-1/(n 2))

Let these four numbers be a1, a2, a3, a4, by the high orange inscription, a2=(a1+a3) 2=16 2=8, (a1, a2, a3 is the difference of the number of the cover of the volt) a3 = 12-a2 = 4, a1 = 16-a3 = 12, a4 = a3 * a3 a2 = 2 (a2, a3, a4 are the equivalent carrying ratio series), so a1, a2, a3, a4 are 12, 8, 4, 2 respectively

You wrote a1=1, right?

a32=1+2（32-1）=63

sn=b1[1-q^(n-1)]/1-q)=(2 -1)*[1-√2 ^(n-1)]/1-√2 )=2 ^(n-1)-1

sn>a32 i.e. leaky jujube 2 (n-1)-1> return to carry and dismantle 63 i.e. 2 (n-1)>64 i.e. hermit 2 [(n-1) 2]>2 6

y=2 x is an increment function.

Therefore (n-1) 2>6 i.e. n>13

That is, the range of n is n>13 and n belongs to n+

Solution: It is known from the person who permeates the mood of the question.

an=a1+(n-1)d=1+2n-2=2n-1sn=a1(a-q (n-1)) (1-q)=2 (n2)-1>a32=63

i.e. 2 ((n-1) 2)>64

n>13

f(an)-f[a(n-1)]=a(n+1)-an=k[an-a(n-1)]

then bn=kb(n-1).

k≠0 so bn is proportional to the group split series.

b1=a2-a1=f(a1)-a

b(n-1)=b1*k^(n-2)

Rule. a2-a1=b1

a3-a2=b2

Initiation....Quietly or suddenly.

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

Add. an-a1=b1+……b(n-1)=b1*[1-k^(n-1)]/1-k)

an=a+[f(a1)-a][1-k^(n-1)]/1-k)

a(n+1) -an 2 + a(n-1) = 0, i.e. a(n+1) + a(n-1) = an 2

In the equal difference series, a(n+1)+a(n-1)=2an, so an 2=2an

and an is not equal to 0Divide both sides of the above equation by an to get an=2 (n is greater than or equal to Chang Xi Brigade 2), which is obviously a constant stool number Lu stupid column. An constant is equal to 2

So s(2k-1)=2*(2k-1)=46, then k=12