Find the maximum value g (a) of the function y = - cosx ^ 2 + acosx + (5 / 8) A-1 / 2 (x is greater than or equal to 0, less than or equal to half)

Find the maximum value g (a) of the function y = - cosx ^ 2 + acosx + (5 / 8) A-1 / 2 (x is greater than or equal to 0, less than or equal to half)

y=-cos^2x+acosx+(5/8)a-1/2=-(cosx-a/2)^2+a^2/4+5a/8-1/2=-(cosx-a/2)^2+(a/2+5/4)^2-57/64
When a > 2, G (a) = 13A / 8-3 / 2, then cosx = 1, x = 0
When - 2

If {an} {BN} is an equal ratio sequence, then {lgan} is an equal difference and {aan} {1 / an} {anbn} {an / BN} is an equal ratio, then what is the tolerance or common ratio?

Let {an} be p, an > 0, P > 0. Let {BN} be q, BN ≠ 0, Q ≠ 0
Sequence {lgan} tolerance = LG (an) - LGA (n-1) = LG [an / a (n-1)] = LG (P)
Sequence {aan} common ratio = AA (n-1) P / a (n-1) = AP
Sequence {1 / an} common ratio = [1 / a (n-1) P] / [1 / a (n-1)] = 1 / P
Sequence {anbn} common ratio = a (n-1) P B (n-1) Q / [a (n-1) B (n-1)] = PQ
Sequence {an / BN} common ratio = [a (n-1) P / b (n-1) q] / [a (n-1) / b (n-1)] = P / Q

The first term of the arithmetic sequence {an} is A1, the tolerance is D, and the sum of the first n terms is SN;
The first term of the arithmetic sequence {an} is A1, the tolerance is D, and the sum of the first n terms is SN. The following four statements are given: ① the sequence {(1 / 2) an} is an equal ratio sequence; ② if A2 + A12 = 2, then S13 = 13; ③ Sn = n * an Nd (n-1) / 2; ④ if d > 0, then Sn must have a maximum value
The correct serial number is_______ Write all the correct serial numbers

It is known that the first term of the arithmetic sequence {an} is A1, the tolerance is D, the sum of the first n terms is Sn, an = a1 + (n-1) d, Sn = n (a1 + an) / 2
① The sequence {(1 / 2) ＾ an} is equal ratio sequence;
(1/2)＾an／(1/2)＾a（n－1）＝（1／2）＾［an－a（n－1）］＝（1／2）＾d
The ratio of the last term to the first term of a sequence is constant, so the sequence is an equal ratio sequence
② If A2 + A12 = 2, then S13 = 13;
If n + M = P + Q, then an + am = AP + aq;
So A2 + A12 = A7 + A7 = 2a7 = 2, the solution is A7 = 1
S13
＝a1＋a2＋．．．＋a12＋a13
＝（a1＋a13）＋（a2＋a12）＋．．．＋（a6＋a8）＋a7
＝13a7
= 13; yes
③Sn=n ×an-nd(n-1)/2；
It is known that an = a1 + (n-1) d, so A1 = an - (n-1) d
sn＝n（a1＋an）／2
＝n［an＋an－（n－1）d］／2
＝n［2an－（n－1）d］／2
= n × an Nd (n-1) / 2; right
④ If d > 0, there must be a maximum value of Sn
If d > 0, then the sequence an is an increasing sequence, the larger n is, the larger an is, and the larger Sn = n (a1 + an) / 2 is, so Sn has no maximum
The correct serial number is (1) (2) (3)
Hope to be able to help you, if you have any questions, please ask, help learning progress!

If the sequence {an} satisfies an = QN (Q > 0, n ∈ n *), then______ (1) the arithmetic sequence is lgan {2}

Because Q ＞ 0, the sequence an = QN (Q ＞ 0, n ∈ n *) is equal ratio sequence, and the common ratio is Q; (2) 1An = 1qn = (1q) n, so it is an equal ratio sequence, and the common ratio is 1q. So (2) is correct. (3) because lgan = lgqn = nlgq, so lgan is an equal difference sequence, and the tolerance is lgq, so (3) is correct. (4) because lga2n = 2lgan = 2lgqn = (2lg ⁡ q) ⋅ n, {lgan2} is an equal difference sequence

What is the number of the power of 2008 in 2006 + the power of 2008 in 2003

Just look at one seat
The first power of 6 is 6, the second power is 6, the third power is 6
Any bit of 6 is 6
So the 2008 power of 2006 is 6
The first power of 3 is 3, the second power is 9, the third power is 7, the fourth power is 1, and the fifth power is 3,
So the 2008 power of 2003 is 1
Therefore, the number of the 2008 power of 2006 + the 2008 power of 2003 is 6 + 1 = 7

Let's relax and make a number game. The first step: take a natural number N1 = 5, calculate nl2 + 1, and record the result as A1. The second step: calculate A1
The first step: take a natural number N1 = 5, calculate nl2 + 1, and record the result as A1;
Step 2: calculate the sum of all the numbers of A1 to get N2, calculate N2 ^ 2 + 1, and the result is A2;
Step 3: calculate the sum of the numbers of A2 to get N3, and then calculate N3 ^ 2 + 1, the result is A3;
…
And so on, then A10 =?

From the meaning of the title
n1=5,a1=26
n2=8,a2=65
n3=11,a3=122
n4=5,a4=26
.
It is found that an has three cycles, so A10 = A1 = 26
If you don't understand, you can ask. If you are satisfied, please accept ~^_^

Let's relax and make a number game. The first step: take a natural number N1 = 5, calculate nl2 + 1, and record the result as A1. The second step: calculate A1
The first step: take a natural number N1 = 5, calculate nl2 + 1, and record the result as A1;
Step 2: calculate the sum of all the numbers of A1 to get N2, calculate N2 ^ 2 + 1, and the result is A2;
Step 3: calculate the sum of the numbers of A2 to get N3, and then calculate N3 ^ 2 + 1, the result is A3;
…
And so on, then A28 =?

Let N1 = 5, then: A1 = N1 & # 178; + 1 = 25 + 1 = 26
If N2 = 2 + 6 = 8, then A2 = N2 & # 178; + 1 = 64 + 1 = 65
N3 = 6 + 5 = 11, then A3 = N2 & # 178; + 1 = 121 + 1 = 122
n4=1+2+2=5,a4=n4²+1=25+1=26
...
Therefore, it can be seen that the number in this column is in the order of 26, 65122, which are arranged repeatedly
And 28 △ 3 = 9... 1
So A28 = 26

Let's relax and do a math game: Step 1: take a natural number N1 = 5, calculate N12 + 1 to get A1; step 2: calculate the numbers of A1
N 2 is obtained by the sum of N 22 + 1, and a 2 is obtained by calculating n 22 + 1;
Step 3: calculate the sum of the numbers of A2 to get N3, and calculate N32 + 1 to get A3;
…
And so on, A2012=

n1=5 a1=5²+1=25+1=26 n2=2+6=8a2=8²+1=64+1=65 n3=6+5=11a3=11²+1=121+1=122 n4=1+2+2=5a4=5²+1=25+1=26 …… Now, when N4 = 5, it returns to N1 state, and it will continue to cycle like this in the future. It can be seen that every three times is a cycle; 2012 / 3 = 6

Let's relax and make a number game: Step 1: take a natural number N1 = 4, calculate N12 + 1 to get A1; step 2: calculate the sum of all the numbers of A1 to get N2, calculate N22 + 1 to get A2; step 3: calculate the sum of all the numbers of A2 to get N3, calculate N32 + 1 to get A3 And so on, the value of a2013 is

According to the meaning of the title:
n1=5,a1=5×5+1=26；
n2=8,a2=8×8+1=65；
n3=11,a3=11×11+1=122；
n4=5,a4=5×5+1=26；
…
∵ three number cycles
Ψ n2013 is the third of the 671st cycle,
∴a2013=a3=122

Let's relax and play a number game: the first step: take a natural number N1 (1 represents the number 1 of n) = 5, and calculate N12 (12 as above)

n12=16