In the equal ratio sequence {an}, the first term A1 ＜ 0, if the sequence {an} has an + 1 ＞ an for any positive integer n, then the common ratio Q should satisfy A.q＞1 B.0＜q＜1 C.1/2＜q＜1 D.-1＜q＜0

In the equal ratio sequence {an}, the first term A1 ＜ 0, if the sequence {an} has an + 1 ＞ an for any positive integer n, then the common ratio Q should satisfy A.q＞1 B.0＜q＜1 C.1/2＜q＜1 D.-1＜q＜0

Choose B
When n = 1
a2=a1q>a1
Namely
a1q-a1>0
a1*(q-1)>0
A10
therefore
q^(n-1)>0
Because n is any natural number
therefore
Q>0
To sum up, the answer is B, 0
Choose B
Let the ratio be q and Q ≠ 0
an=a1*q^(n-1)
an+1=a1*q^n
From the topic meaning A1 * q ^ n > A1 * q ^ (n-1)
That is A1 * q ^ (n-1) * (Q-1) > 0
A1
There is a column number A1 = 1, after A2, A3, A4 The rule is as follows: if An-2 is a natural number and has not been written before, then write an + 1 = An-2,
There is a column number A1 = 1, after A2, A3, A4 The rules are as follows:
If An-2 is a natural number and has not been written before, then write an + 1 = An-2, otherwise write an + 1 = an + 3, and then calculate the value of A6
（A2）=（A1）3 = 1 +3 = 4
（A3）=（A2）+3 = 4 +3 = 7
（A4）=（A3）-2 = 7 - 2 = 5
（A5）=（A4）-2 = 5-2 = 3
（A6）=（A5）+3 = 3 +3 = 6
No
To understand, please send future questions, preferably in picture format, or in parentheses, or useful math questions ambiguous!
Your problem should be corrected
If a (n-2) is a natural number and it is not written before, you write a (n + 1) = a (n) - 2, otherwise write a (n + 1) = a (n) + 3
(A2)=(A1)+3=1+3=4
(A3)=(A2)+3=4+3=7
(A4)=(A3)-2=7-2=5
(A5)=(A4)-2=5-2=3
(A6)=(A5)+3=3+3=6
If you don't understand, please ask
In addition, if you want to ask questions later, you'd better send the picture format or add brackets, otherwise the math problem is very ambiguous!
Your question should be corrected to
If a (n-2) is a natural number and has not been written before, then write a (n + 1
(A2)=(A1)+3=1+3=4
(A3)=(A2)+3=4+3=7
(A4)=(A3)-2=7-2=5
(A5)=(A4)-2=5-2=3
(A6)=(A5)+3=3+3=6
If you don't understand, please ask
In addition, if you want to ask questions later, you'd better send the picture format or add brackets, otherwise the math problem is very ambiguous!
Your question should be corrected to
If a (n-2) is a natural number and has not been written before, then write a (n + 1) = a (n) - 2, otherwise write a (n + 1) = a (n) + 3
Given that the perimeter of ABC is 6, the modules of BC vector, CA vector and ab vector are a, B and C in turn. Prove 0 in an equal ratio sequence
Let | vector BC | = a, | vector Ca | = B, | vector ab | = C, then there are: a + B + C = 6, B ^ 2 = AC | a + C = 6-b, AC = B ^ 2, so a and C are two real number roots of the equation x ^ 2 - (6-b) x + B ^ 2 = 0. According to WIDA's theorem: (6-b) ^ 2-4b ^ 2 ≥ 036-12b + B ^ 2-4b ^ 2 ≥ 0b ^ 2 + 4b-12 ≤ 0 (B + 6)
Let | vector BC | = a, | vector Ca | = B, | vector ab | = C, then:
a＋b＋c＝6，b^2＝ac
∴a＋c＝6－b，ac＝b^2
So a and C are the two real roots of the equation x ^ 2 - (6-b) x + B ^ 2 = 0
According to Weida's theorem:
(6－b)^2－4b^2≥0
36－12b＋b^2－4b^2≥0
b^2＋4b－12≤0
(b＋6)(b－2)≤0
Due to B ＞... Unfold
Let | vector BC | = a, | vector Ca | = B, | vector ab | = C, then:
a＋b＋c＝6，b^2＝ac
∴a＋c＝6－b，ac＝b^2
So a and C are the two real roots of the equation x ^ 2 - (6-b) x + B ^ 2 = 0
According to Weida's theorem:
(6－b)^2－4b^2≥0
36－12b＋b^2－4b^2≥0
b^2＋4b－12≤0
(b＋6)(b－2)≤0
Because b > 0, B ≤ 2
On the other hand, | a-c | B
∴(a－c)^2＜b^2
(a＋c)^2－4ac＜b^2
(6－b)^2－4b^2＜b^2
b^2＋3b－9＞0
From B ＞ 0, we know: B ＞ (- 3 + 3 √ 5) / 2
∴(－3＋3√5)/2＜b≤2
And vector Ba, vector BC
= | vector BA ·| vector BC · · CoSb
＝ac·(a^2＋c^2－b^2)/(2ac)
＝(a^2＋c^2－b^2)/2
＝[(a＋c)^2－2ac－b^2]/2
＝[(6－b)^2－3b^2]/2
＝－(b＋3)^2＋27
From (- 3 + 3 √ 5) / 2 ＜ B ≤ 2: (3 + 3 √ 5) / 2 ＜ B + 3 ≤ 5
(27＋9√5)/2＜(b＋3)^2≤25
－(27＋9√5)/2＞－(b＋3)^2≥－25
－(27＋9√5)/2＋27＞－(b＋3)^2＋27≥－25＋27
That is: (27-9 √ 5) / 2 ＞ - (B + 3) ^ 2 + 27 ≥ 2
The value range of Ba and BC is: [2, (27-9 √ 5) / 2]
Finding discontinuities and their types: F (x) = (x ^ 2-1) / (x ^ 2-3x + 2)
I just came into contact with calculus, can you tell me in detail the process of judging the number and type of discontinuities
This question is wrong. There is no break point in this question. I have answered all the other questions in your other question
The common ratio of {an} is Q, the product of the first n terms is TN, and satisfies A1 > 1, a99a100 > 1, a99-1 / a100-1
,a99-1/a100-11,
1. So A99 > 1, A100
According to (a99-1) / (a100-1) 1, A100
There is a column of numbers, A2, A3, A4 If A1 = 2, then A2009 is
1 B 2 C 2/1 D -1
a1 = 2
a2 = 1 - 1/2 = 1/2
a3 = 1- 2 = -1
a4 = 1 - (-1) =2
Three numbers one cycle
2009 % 3 = 2
So A2009 = 1 / 2
Choose C
It is known that in the triangle ABC, the height of a (2,1), B (3,2), C (- 3, - 1), BC side is ad, and the coordinates of point D and vector AD are obtained
Let D (x, y) vector BC = (- 3, - 6) vector ad = (X-2, Y-1) because vector BC is perpendicular to vector ad, so they multiply by each other = 0, so - 3 * (X-2) - 6 * (Y-1) = 0. ① and point D on line BC, the equation of straight line BC is y = (- 189; X + (- 189); ② simultaneous ① and ② two squares
F (x) = X-1 / x ^ 2 + X-2
f(x)=(x-1)/(x-1)(x+2),
When x = 1, x = - 2, the function has no meaning, so it is the breakpoint between functions,
They all belong to the second kind of discontinuities,
If Lim [x → 1] f (x) = 1 / 3, the limit exists. If f (1) = 1 / 3, then x = 1 is a function removable discontinuity
Let {an} be a sequence of positive numbers, the sum of the first n terms is Sn, and for all natural numbers n, the sum of an and 1 is equal
The general term formula of sequence {an} can be obtained when the term is equal to the middle term of Sn in 1
From the meaning of the title
(an +1)/2=√(Sn×1)
Sn=[(an +1)/2]²
When n = 1, S1 = A1 = [(a1 + 1) / 2] &
(a1-1)²=0
a1=1
When n ≥ 2,
Sn=[(an +1)/2]² S(n-1)=[(a(n-1) +1)/2]²
Sn-S(n-1)=an=[(an+1)/2]²-[(a(n-1)+1)/2]²
4an=an²+2an+1-[a(n-1)+1]²
(an -1)²-[a(n-1)+1]²=0
[an-1+a(n-1)+1][an -1-a(n-1)-1]=0
[an+a(n-1)][an-a(n-1)-2]=0
The sequence is positive, an + a (n-1) > 0. If the equation holds, only an-a (n-1) = 2 is the fixed value
Sequence {an} is an arithmetic sequence with 1 as the first term and 2 as the tolerance
an=1+2(n-1)=2n-1
The general formula of sequence {an} is an = 2N-1
(1) [^ 2] = (SN) / 2] = 1
When n = 1, A1 = S1 = [(1 + A1) ^ 2] / 4, so (1-a1) ^ 2 = 0, so A1 = 1
When n ≥ 2, s (n-1) = {[1 + a (n-1)] ^ 2} / 4
① - 2, SN-S (n-1) = [(1 + an) ^ 2] / 4 - {[1 + a (n-1)] ^ 2} / 4
And SN-S (n-1) =... Expansion
It is known from the meaning of the title that: [(1 + an) / 2] ^ 2 = 1 × Sn, that is: SN = [(1 + an) ^ 2] / 4
When n = 1, A1 = S1 = [(1 + A1) ^ 2] / 4, so (1-a1) ^ 2 = 0, so A1 = 1
When n ≥ 2, s (n-1) = {[1 + a (n-1)] ^ 2} / 4
① - 2, SN-S (n-1) = [(1 + an) ^ 2] / 4 - {[1 + a (n-1)] ^ 2} / 4
And SN-S (n-1) = an, so [(1 + an) ^ 2] / 4 - {[1 + a (n-1)] ^ 2} / 4 = an,
It is reduced to [an + a (n-1)] × [an-a (n-1) - 2] = 0
Because the sequence {an} is composed of positive numbers, that is, an > 0, a (n-1) > 0
So an + a (n-1) > 0, so an-a (n-1) - 2 = 0, then an-a (n-1) = 2, which is a constant
So the sequence an is an arithmetic sequence with 1 as the first term and 2 as the tolerance
The general formula of an is an = 1 + 2 (n-1) = 2N-1 (n ∈ n +)
(1)[a(n) 2]^2=8s(n),
[a(1) 2]^2=8s(1)=8a(1),[a(1)-2]^2=0,a(1)=2.
When [a (2) 2] ^ 2 = 8s (2) = 8 [a (1) a (2)], [a (2) - 2] ^ 2 = 8A (1) = 16, a (2) > = 2, a (2) - 2 = 4, a (2) = 6,
Zero
If A1 = 1, A2 = 4, A3 = 7, A4 = 10, an = 31, then n=
N=1
According to the situation of A1 to A4, it is judged that this is an arithmetic sequence with tolerance d = 3
From the n-th formula of the arithmetic sequence an = a1 + (n-1) d, the above numbers are brought in to obtain n = 11