If a1 + A4 = 18, A2 + a3 = 12, then the sum of the first eight terms of the sequence is () A. 513B. 512C. 510D. 2258

If a1 + A4 = 18, A2 + a3 = 12, then the sum of the first eight terms of the sequence is () A. 513B. 512C. 510D. 2258

Let the first term of the equal ratio sequence be A1, the common ratio be q ∵ a1 + A4 = 18, A2 + a3 = 12 ∵ A1 (1 + Q3) = 18a1q (1 + Q) = 12. By dividing the two formulas, 2q2-5q + 2 = 0 can be obtained by taking the common ratio Q as an integer, q = 2, A1 = 2 can be obtained by substituting it into the sum formula of the equal ratio sequence, S8 = 2 (1 − 28) 1 − 2 = 510
The square of the complex Z is - 7. What is Z?
z=a+ib
z^2=a^2-b^2+2iab=-7+0i
So AB = 0
a^2-b^2=-7
So a = 0, B = plus or minus (root 7)
So z = ± (root 7) I
There is no exact number, just as the square of I equals - 1. I has no exact number.
Let a = {X! - 1
a>=-1
It is known that {an} is a q-equal ratio sequence with common ratio, and A1, A3 and A2 are equal difference sequence. In 1, we find the value of Q, and in 2, let {BN} be an equal difference sequence with 2 as the first term and Q as the tolerance, the first n term and Sn
The complex 2i1 − I equals ()
A. -1+iB. -1-iC. 1+iD. 1-i
The complex number 2i1 − I = 2I (1 + I) (1 − I) (1 + I) = 1 + I
The solution set of ax square + BX + 1 greater than 0 is {x | x is not equal to negative 2} to find the value of a and B
Ax ^ 2 + BX + 1 ＞ 0 according to the meaning of the problem, the function f (x) = ax ^ 2 + BX + 1 has the opening upward and the vertex x coordinate is x = - 2, so a > 0, the equation AX ^ 2 + BX + 1 = 0 has two equal real roots, which is x = - 2 △ = B ^ 2-4a = 0,4a = B ^ 2. Substituting x = - 2 into the equation AX ^ 2 + BX + 1 = 0, we get: 4a-2b + 1 = 0b ^ 2-2b + 1 = 0 (B-1) ^ 2 = 0b = 14a = 1, a = 1 / 4
The first term of the arithmetic sequence {an} is A1 = 1, and the common ratio Q ≠ 1. If A1, A2 and A3 are the first, second and fifth terms of the arithmetic sequence in turn, find Q
Because A1, A2 and A3 are the first, second and fifth terms of the arithmetic sequence in turn
So a3-a1 = 4 (a2-a1)
Common ratio Q ≠ 1
Q = 3
What is 2I of 1-I?
2I of 1-I
=2i(1+i)/[(1-i)(1+i)]
=(2i-2)/2
=-1+i
I-1
If the numerator and denominator are multiplied by 1 + I at the same time, I-1 will be obtained
Multiply (1 + I) to get 2I (1 + I) / (1-I) (1 + I) = (2i-2) / 2 = - 1 + I
Given the set a = {the square of X + X-2 is less than or equal to 0}, B = {x | 2 is less than x + 1 is less than or equal to 4}, let the set C = {the square of X + BX + C is greater than 0}, and satisfy (AUB) intersection C = empty set, (AUB) UC = R, find the value of B and C
First, find the interval of A
X = (- B ± √ (b ^ 2-4ac)) / 2a, a = [- 2,1]
Similarly, B = [1,3]
A and B = [- 2,3], so C = (x3)
(-b-√(b^2-4ac))/2a3
Where a = 1
Solving binary quadratic equations
B = - 13, C = 30
It may be wrong, but that's the way
Above, B = - 1, C = - 6
In the equal ratio sequence {an}, A1. A2. A3 = 27, A2 + A4 = 30, try to find the sum of A1 and common ratio, and then find the sum of the first six terms S6
For the sequence of equal ratio, there is (A2) ^ 2 = a1a3. Combining with the conditions, A2 = 3, A4 = 27. So q = 3, A1 = 1, S6 = 364