Factorization factor (1 / 2) a ^ 3b-2a ^ 2B ^ 2 + 2Ab ^ 3

Factorization factor (1 / 2) a ^ 3b-2a ^ 2B ^ 2 + 2Ab ^ 3

Factorization of (2a-3b) 2 - (a + 2b) 2

There are several numbers, denoted as A1, A2, A3,... If A1 = - 1 / 2, from the second number, each number is
If A1 = - 1 / 2, starting from the second number, each number is equal to the reciprocal of the difference between 1 and the number in front of it, then A2 =? A2009 =?
It's better to say why~

Observation sequence:
a1=-1/2;
a2=1/[1+1/2]=2/3;
a3=1/[1-2/3]=3;
a4=1/[1-3]=-1/2;
a5=2/3;
a6=3;
.
Since the calculation method is the same, we will continue to repeat - 1 / 2,2 / 3,3, so the 3N term of this sequence is 3;
Therefore, a2007 = 3, then A2009 = 2 / 3

There are several numbers, the first number is A1, the second number is A2, the third number is A3,..., and the nth number is an, if A1 = - 1 / 2,
From the second number, each number is equal to the reciprocal of "1" and the number before it. It is calculated as: A1 = A2 = A3 = A4 = A2003 = a2004 = a2005=
It's urgent! Come on!
Each number is equal to the reciprocal of the difference between "1" and the number before it.

The first number is denoted as A1, and the nth number is denoted as an, if A1 = - 1 / 2, from the second number, every number is equal to the reciprocal of the difference between 1 and the number in front of it, A1 = - 1 / 2, A2 = 1 / [1 + 1 / 2] = 2 / 3, A3 = 1 / [1-2 / 3] = 3, A4 = 1 / [1-3] = - 1 / 2... It can be seen that every three are a cycle. 2003 / 3 = 667... 2, so, A2003 = A2 = 2 / 3

Given the set a = {x | x = A0 + A1 × 3 + A2 × 32 + a3 × 33}, where AK ∈ {0, 1, 2} (k = 0, 1, 2, 3), and A3 ≠ 0, then the sum of all elements in a is equal to ()
A. 3240B. 3120C. 2997D. 2889

It can be seen from the meaning of the question that A0, A1, A2 each have three kinds of methods (all can take 0, 1, 2), A3 has two kinds of methods, according to the step-by-step counting principle, there are a total of 3 × 3 × 3 × 2 methods, when A0 takes 0, 1, 2, A1, A2 each have three kinds of methods, A3 has two kinds of methods, there are a total of 3 × 3 × 2 = 18 methods, that is, the sum of all numbers containing A0 term in set a is (0 + 1 + 2) × 18; similarly, all numbers containing A1 term in set a can be obtained The sum of several numbers is (3 × 0 + 3 × 1 + 3 × 2) × 18; the sum of all numbers with A2 in set a is (32 × 0 + 32 × 1 + 32 × 2) × 18; the sum of all numbers with A3 in set a is (33 × 1 + 33 × 2) × 27; according to the principle of classification and counting, the sum of all elements in set a is s = (0 + 1 + 2) × 18 + (3 × 0 + 3 × 1 + 3 × 2) × 18 + (32 × 0 + 32 × 1 + 32 × 2) × 18 + (33 × 1 + 33 × 2) × 27 = 18 (3 + 9 + 27) + 81 × 27 =702 + 2187 = 2889

In the arithmetic sequence {an}, the first term A1 = 0, tolerance D ≠ 0, if AK = a1 + A2 + a3 + +A7, then K=______ ．

From A1 = 0, tolerance D ≠ 0, we get an = (n-1) d, then AK = a1 + A2 + a3 + +A7 = (a1 + A7) + (A2 + A6) + (A3 + A5) + A4 = 7a4 = 21d, and AK = (k-1) d, so k-1 = 21, the solution is k = 22

If all the prime factors of 510510 are A1, A2... AK in descending order, find (a1-a2) (A2-A3). (ak-1-ak)
In seven days, please

No programming
510510=2*3*5*7*11*13*17
(A1-A2)(A2-A3).(Ak-1-AK)
=(-1)*(-2)*(-2)*(-4)*(-2)*(-4)
=128

There are several numbers, denoted as A1, A2, A3 An, if A1 = - 1 / 2, from the second number, each number is equal to the sum of 1 and the number before it
There are several numbers, the first number is A1, the second number is A2 If A1 = 1 / 2, starting from the second number, each number is equal to "the reciprocal of the difference between 1 and the number in front of it". Try to calculate: A2 = A3 = A4 = A5 = please calculate the values of A2012 and a2016

It is known that A1 = 1 / 2, A2 = 2, A3 = - 1, A4 = 1 / 2, A5 = 2
According to the law, every three cycles, A2012 = 2, a2016 = - 1
This kind of sequence is called swing sequence

There are several numbers, A1, A2, A3. An. If A1 = - 1 / 2, from the second number, each number is equal to "1" and the number before it
Then the top: is the last
Q: 1 A2 = A3=
2 find the value of A9; A10; a11 (note the sign)
3 is there a value of m such that M / (an-1; an; an + 1) = 1? If so, the value of M is requested

(1) A2 = 1 / (1 - (- 1 / 2)) = 2 / 3, A3 = 1 / (1-2 / 3) = 3 (2) A4 = 1 / (1-3) = - 1 / 2, so A9 = A3 = 3, A10 = A1 = - 1 / 2, a11 = A2 = 2 / 3 (3) A1 = - 1 / 2An = 1 / (1-A [n-1]) a [n-1] = (an-1) / anan-1 * an * an + 1 = (an-1) / an * an * 1 / (1-an) = - 1m / (an-1 * an * an + 1) = - M = a

Why A3 * A4 = A1 * A6 = A2 * A5
Why is it in the sequence of equal numbers
a3*a4=a1*a6=a2*a5
Who can help me to explain this problem
In the equal ratio sequence, given that A4 * A3 = - 128, A3 + A6 = 28, and the common ratio is an integer, find the value of A3 + A4 + A5 + A6

Because if you multiply with the power of the base, the base will not change and the index will add,
So A3 * A4 = A7, A1 * A6 = A7, A2 * A5 = A7