High school three number series questions! Okay, you can add points, thank you

Let the first term be A1 and the tolerance be D

It can be obtained from the formula an=a1+(n-1)d.

a1+a1+d+a1+2d=21

6a1+15d=24

The solution yields a1=9 and d=-2

So the general formula is an=11-2n

Order an>0

Get 11 2n>0

n> i.e. the first 5 terms are all positive, and from the 6th term onwards are negative.

Let the first term be A1 and the tolerance be D

s=a1*n+n(n-1)d 2, then 21=3a1+3d, 24=6a1+15d, the solution is a1=9, d=-2

So the general formula is an=11-2n

There are only 9 items, which are 11, 9, 7, 5, 3, 1, 1, 3, 5, which adds up to 65

The sum of the first three terms should be 3 first terms plus 3 tolerances.

The sum of the last three terms should be 3 first terms, plus 3 + 4 + 5 = 12 tolerances = 24-21 = 3 minus 9 tolerances = -18

With a tolerance of -2, you should be able to do it yourself.

Answer: For this question, it is possible to consider doing it positively.

It will cause us unnecessary trouble, so we should think about the problem in a different way. Mathematical ideas are broad and profound. If this road is not passable, the key will collapse, and Xiang will go another way.

Therefore, when we do this kind of question, we should deal with it according to his opposing problem.

Step 1: Find a situation where there is no pair, let's set this as event p

Here's an analysis of how to answer this event.

In the chapter on probability, we know that finding a probability requires the total basic event, and the total event under that condition. It has 10 pairs of shoes, 20 and only 8 of them.

For this question, the total basic event is [c20

8] Shecong 20

Only 8 of them are selected, and I will put a comma between these two numbers below. Please forgive me for not being able to play out due to the bad clothes), and so on.

Choose one of the ten pairs of shoes [C10,1], and then choose any one of these shoes to come out as [C2,1].

Then choose one of the remaining 9 pairs of shoes as [C9,1], and then choose any one of these shoes as [C2,1].

Then choose any pair of the remaining 8 pairs of shoes as [C8,1]. Then choose any one of these shoes as [C2,1].

And so on until you choose one of the remaining 3 pairs of shoes [C3,1], and then choose any of these shoes as [C2,1].

then there is p= [c20,8].

So the probability of at least one pair is p(find)=1-p

There are questions in the question"At least"Two words.

Obviously, the reverse side is used here.

The probability of not having a pair is (8 pairs of brothers are selected.)

Choose one of each).

p=(10*9/2*2^8)/(20!/(12!*8!The answer to this question is (1-p).

a3=a1*q 2=e (b2)=e 18a6=a1*q 5=e (b6)=e 12 then: a6 a3=q 3=e 12 e 18=e (-6) yields: q=e (-2), a1=e 22

General term formula for proportional series:

an=e^(24-2n)

The sequence satisfies bn=ln(an).

General term formula for the series:

bn=24-2n

When n=12, bn=0

an≠1 does not have bn=0, n≠12

The maximum value of the sum of the first n terms of bn}:

b1+b2+……b11

b1+b2+……b12

Solution: 1Repeat substitution to find out.

2.In the case of proof, the process is a bit cumbersome, similar to the Lasong answer to the first question, find a(n+1) a(n+2)...a(n+4) can be seen [all are represented by a(n)].

a(n+4)=a(n), so that it is a series of periods with period t=4.

3.The sum of every 4 items is the same, so s(2008)=502*s(4)=2008, this is an equation about a, it should be able to be solved, and then check the handwriting to prove whether the wheel potato Zhengxia satisfies 1 a 4.

an+1+an=2n-44 So a(n+2)+a(n+1)=2(n+1)-44 subtract these two equations to get a(n+2)-an=2

So even terms are divided into a series of equal differences, and odd terms are formed into a series of equal differences.

Since a1=-23 a2=-19, then a(2n-1)=2n-25 a2n=2n-21

So an=n-24 (n is odd) n=n-21 (n is even) 2) when n is odd.

sn=(a1+a3+,,an)+(a2+a4+,,an-1)=[23+(-21)+、n-24]+[19+(-17)+、n-23]=(2n, 2-89n-5) 4 Find its minimum value, i.e., when n=22, get it and calculate it yourself.

When you're an even number, you can talk about it in the same way and come up with a minimum.

Compare these two minima, which is the smaller one, and that's it.

(1) The equivalence of the problem to the condition is [a(n+1)-(n+1-45 2)]=an-(n-45 2)] by the matching method

So {an-(n-45 2)} is a proportional sequence of q=-1, and the first term is -23-1+45 2=-3 2

So an=(-3 2)*(1) (n-1)+n-45 2

1. s(n)=2n 2+1, then s(n-1)=2(n-1) 2+1; a(n)=s(n)-s(n-1);(n)=2(2n-1);

2. Same as above, proportional series q=-1 3,a(1)=s(1); a(n)=4 3*(-1 3) (n-1);

3. Same as above, a(n)=s(n-1),(n>=2),a(n-1)=s(n-2), proportional series, q=2;a(n)=3*(2) (n-1);

4. From the known conditions, the first two terms are combined to obtain (1 2)*(1+n) 2, (n is an odd number), (1 2)*n 2+1 4, (n is an even number);

After unification, we get: (1 4)*(n+1);

5. What is the meaning of the question s, I didn't understand - -

6. Does sqr mean to seek roots? Rationalization, summing, intermediate term cutting, s(n)=-sqrt(1)+sqrt(n+1)=10,n=120;

7. Equal difference + equal ratio; a(n)=(2n-1)+(1/2)^n,s(n)=n^2+1-(1/2)^n;

1，sn=2n^2+1

an=sn-s(n-1)=4n-1

2, s(n+1)=1+a(n+1)/4a(n+1)=s(n+1)-sn=(a(n+1)-an)/43a(n+1)=4an

a(n+1)/an=4/3

an=(4/3)^(n-1)xa1

a1=1+a1/4 a1=4/3

an=(4/3)^n

3,an=a1+a2+…+an-1

a1+a2+…+an-1+an=2(a1+a2+…+an-1)sn=2s(n-1)

sn=2^(n-1)s1=3x2^(n-1)s-s(n-1)=an=3x2^(n-2)..n>=2a1=3...n=1

4,f(0)+f(1/n)+f(2/n)+.f[(n-1)/n]+f(1)=1/2((f0+f1)+(f1/n+f(n-1)/n)(f2/n+f(n-2)/n)..

1/2xn(1/2)=n/4

5,s3=s11

a1+a2+a3=a1+a2+a3+a4+a...a11a11+a10+a9+..a4=0

8a1+52b=0

2a1+13b=0

a1+6b+a1+7b=0

a7+a8=0

a1>0

suoyia7>0

a8<0

Top 7 max

6, What is SPR?

7，sn=1+3+5+7+。。2n-1） +1/2+1/4+1/8+。。1/2^n=n^2+1-1/2^n