Let 1 = A1 ≤ A2 ≤ Where a1, A3, A5 and A7 are equal ratio sequences with common ratio Q and A2, A4 and A6 are equal difference sequences with tolerance 1, then the minimum value of Q is () A. 33B. 1C. 3D. 3

Let 1 = A1 ≤ A2 ≤ Where a1, A3, A5 and A7 are equal ratio sequences with common ratio Q and A2, A4 and A6 are equal difference sequences with tolerance 1, then the minimum value of Q is () A. 33B. 1C. 3D. 3

∵1=a1≤a2≤… The minimum value of A6 is 3, the minimum value of A7 is 3, A1 = 1, and A1, A3, A5 and A7 are equal ratio sequences with common ratio Q. there must be Q > 0, a7 = a1q3 ≥ 3, Q3 ≥ 3, and the minimum value of Q is 33
The following numbers 2, 3 and 6 are required to be subtracted first and then divided. How can they be equal to zero
2—6/3
100^2-99^2+98^2-97^2+96^2. +2^2-1
Why is 202 × 50 divided by 2 in the last step?
100^2-99^2+98^2-97^2+96^2.+2^2-1= (100²-99²) + (98²-97²) + (96²-95²) + .+ (2²-1)= (100+99)(100-99) + (98+97)(98-97) + (96+95)(96-95) + .+ (2+1)(2-1)= 199 + 195 + 191...
100^2-99^2+98^2-97^2+96^2...... +2^2-1
=(100+99)(100-99)+(98+97)(98-97)+(96+95)(96-95)+...(2+1)(2-1)
=199+195+194+...+3
=(3+199)*50/2=5050
The last step is the summation formula of arithmetic sequence
The first step is to simplify the formula of square difference to 199 + 195 +... +3
The second step is to sum the arithmetic sequence: (first term + last term) * number of terms divided by two, that is (199 + 3) * 50 / 2
In the positive number sequence {an}, a4a5 = 32, then log2a1 + log2a2 + + log2 a8=
From the meaning of the title,
Original formula = log2 [A1 * A2 *. * A8]
=log2(a4*a5)^4
=log2(2^20)
=20
a1a8=a2a7=a3a6=a4a5=32
Original formula = log2 (a1a2a3a4a5a6a7a8) = log2 (32 ^ 4) = log2 ((2 ^ 5) ^ 4) = 20
a4a5=32=a1^2*Q^7
log2 a1+ log2 a2+…… + log2 a8
=log2 a1*a2*a3*a4*a5*a6*a7*a8
=1og2 a1^8*Q^28
=1og2 32^4
=20
What is the difference between a number minus 116 and the sum of 227 and 313?
227 + 313 + 116 = 11821 + 116 = 9514; a: this number is 9514
Calculation: 100-99 + 98-97 + 96-95 + +2-1．
The original formula = (100-99) + (98-97) + (96-95) + +（2-1）=1+1+… +1=50．
If a1 + A8 = 387, A4 * A5 = 1152, then the general term an of the sequence is
a1+a8=a4+a5=387
a4*a5=1152
A4 = 3 A5 = 384 or A4 = 384 A5 = 3
So an = 3 (128) ^ (n-4) or 384 (128) ^ (4-N)
An = (3 times 2 to the power of n-1) or (384 times 1 / 2 to the power of n-1)
The first item is 3, the ratio is 2 or the first item is 384, the ratio is 1 / 2
A1 + A8 = a1 + A1 * q ^ 7 = 387, that is, A1 * Q7 = 387 - A1
A4 * A5 = A1 * q ^ 3 * A1 * q ^ 4 = 1152, that is, A1 * A1 * q ^ 7 = 1152..... 2
1 style dairu 2
a1 * (387-a1) = 1152
a1^2 - 387*a1 + 1152 = 0
a1=3 , a1=384
Substituting 1, we get
q=2 , q=1/2
∴an = 3 * 2^(n-1) , an = 384 / 2^(n-1)
The sum of three times and five of a number is equal to the difference between the number minus one. What is the number
3x+5=x-1
2x=-6
x=-3
Let this number be a, then:
3xa + 5 = A-1, a = - 3
-3
3X+5=X-1
3X-X=-1-5
2X=-6
X=-3
-99+100-97+98-95+96+.-1+2
Please write down the process
-99+100-97+98-95+96+.-1+2
=（-99+100）+（-97+98）+（-95+96）+.+（-1+2）
=1 + 1 + 1. + 1 has 100 △ 2 = 50 ones
=50
-99+100-97+98-95+96+......-1+2
=（-99+100）+（-97+98）+（-95+96）+......+（-1+2）
=1 + 1 + 1... + 1 has 100 △ 2 = 50 ones
=50
In the equal ratio sequence {an}, if A1 = 128, a8 = 1
(1) Find the common ratio Q and A12; (2) prove: take out the first item, the fourth item and the seventh item in sequence {an} Item 3n-2 The new number of terms {a 3n-2} (n ∈ n *) is still an equal ratio sequence
a8=a1q^7
q^7=1/128
q=1/2
So ^ A1 = 2 * A1 = 8
bn=a(3n-2)
Then B (n + 1) = a (3N + 1)
So B (n + 1) / BN
=a1*q^(3n)/[a1*q^(3n-3)]
=q^3
So BN = a (3n-2) is still an equal ratio sequence