Let the sum of the first n terms of the sequence an be Sn, and the points (n, Sn / N) are all on the image 1 of the function y = 3x-2, then the general term formula of the sequence {an} is obtained Let BN = 3 / ana (n + 1), tn be the sum of the first n terms of the sequence {BN}, Let tn be

Let the sum of the first n terms of the sequence an be Sn, and the points (n, Sn / N) are all on the image 1 of the function y = 3x-2, then the general term formula of the sequence {an} is obtained Let BN = 3 / ana (n + 1), tn be the sum of the first n terms of the sequence {BN}, Let tn be

When Sn / N = 3n-2sn = 3N ^ 2-2nn > 1, an = SN-S (n-1) = 6n-5n = 1, A1 = S1 = 1, so for all N, an = 6n-5bn = 3 / [(6n-5) (6N + 1)] = (1 / 2) [1 / (6n-5) - (6N + 1)] TN = (1 / 2) [1-1 / 7 + 1 / 7-1 / 13 +... - 1 / (6N + 1)] = 3N / (6N + 1) TN = 1 / 2-1 / (12n + 2)
If you take the point in, you will get Sn = 3N * 2-2n, which means it is an arithmetic sequence! Single your department can directly say that it is, as long as you understand it, or the teacher will deduct your points! The first n-1 term and Sn are obtained_ 1=3(n-1)*2-2(n-1)
Then Sn SN_ 1 can get the general formula of an! But remember to verify the first item, and note that n is greater than 1, otherwise the previous n-1 is meaningless! ... unfold
If you take the point in, you will get Sn = 3N * 2-2n, which means it is an arithmetic sequence! Single your department can directly say that it is, as long as you understand it, or the teacher will deduct your points! The first n-1 term and Sn are obtained_ 1=3(n-1)*2-2(n-1)
Then Sn SN_ 1 can get the general formula of an! But remember to verify the first item, and note that n is greater than 1, otherwise the previous n-1 is meaningless! Put it away
By substituting a point into a straight line, we can get: SN = 3N square - 2n... According to an = SN-S (n-1)
The points (n, Sn / N) are all on the image of function y = 3x-2
So Sn / N = 3n-2
That is Sn = 3N ^ 2-2n
When n = 1, A1 = S1 = 1
When n > = 2, an = sn-sn-1
=(3n^2-2n)-[3(n-1)^2-2(n-1)]
=6n-5
The above formula also holds for n = 1
So an = 6n-5
Define the function y = f (x), X ∈ D, if there is a constant C, for any x1 ∈ D, there is a unique x2 ∈ D, such that f (x1) + F (x2) & nbsp; 2 = C, then the mean value of function f (x) on D is C. given that f (x) = lgx, X ∈ [10100], then the mean value of function f (x) = lgx on X ∈ [10100] is ()
A. 32B. 34C. 710D. 10
According to the definition, function y = f (x), X ∈ D, if there is a constant C, for any x1 ∈ D, there is a unique x2 ∈ D, such that f (x1) + F (x2) & nbsp; 2 = C, then the mean value of function f (x) on D is called C. let x1 · x2 = 10 × 100 = 1000, when x1 ∈ [10100], select x2 = 1000x1 ∈ [10100]
The sum of the first n terms of the sequence {an} is Sn, and the points (n, Sn / N) are all on the image of the function y = - x + 9
sn/n=9-n
sn=9n-n²
When n = 1, A1 = s1-9-1 = 8
When n ≥ 2, an = SN-S (n-1) = 9n-n & # 178; - [9 (n-1) - (n-1) &# 178;] = 9n-n & # 178; - (- N & # 178; + 11n-10) = - 2n + 10
N = 1 is also satisfied
So an = - 2n + 10, Sn = 9n-n & # 178;
(n, Sn / N) satisfies the function y = - x + 9
Then s (n) / N = - N + 9
S(n)=-n^2+9n
a(n)=S(n)-S(n-1)=10-2n (n≥2)
A (1) = s (1) = 1 + 9 = - 1 + 9 = 8, satisfying the equation a (n) = 10-2n
So a (n) = 10-2n, s (n) = - n ^ 2 + 9N
Sn/n=-n+9
∴Sn=-n^2+9n
a(n)=S(n)-S(n-1)=10-2n
a(1)=S(1)=8,
a(n)=S(n)-S(n-1)=10-2n
Because (n, Sn / N) is carried in on the function y = - x + 9 to get Sn / N = - N + 9, and Sn = - n ^ 2 + 9N is sorted out from A1 = S1 = - 1 + 9 = 8; an = SN-S (n-1) = - 2n + 10
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If there is a constant K, for any two different real numbers X1 and X2 in the domain D, f (x 1) - f (x 2) holds, for the function f (x) = ㏑ x + 1
If f (x) = ㏑ x + 1 / 2x Λ satisfies Lipschitz condition in the interval [0, ∞], what is the maximum value of constant k
k≥|f(x1)-f(x2)|/|x1-x2|
|f(x1)-f(x2)|/|x1-x2|=1/√x1+√x2
Only the maximum value of 1 / √ X1 + √ X2 is the minimum value of K
Obviously, when X1 = x2 = 1, there is a maximum value of 1 / 2
So the minimum value of K is 1 / 2
The sum of the first n terms of the equal ratio sequence {an} is SN. It always belongs to positive integer for any n. The point (n, Sn) is on the image of the function y = B ^ x + R
(b > 0 and B ≠ 1, B, R are constants)
1. Find the value of R
2. When B = 2, note BN = (n + 1) / 4An (n is a positive integer), and find the first n terms of {BN} and TN
1. (n, Sn) when substituting y = B ^ x + RSN = B ^ n + RN > = 2, an = SN-S (n-1) = B ^ n + R-B ^ (n-1) - r = (B-1) × B ^ (n-1) to make {an} an equal ratio sequence, A1 also needs to satisfy the above formula A1 = S1 = B + r = (B-1) × 1R = - 12. B = 2, an = 2 ^ (n-1) BN = (n + 1) / (4 × an) = (n + 1) / 2 ^ (n + 1) TN = B1 + B2 + B3 + +Bn=2/2...
If f (x) = a ^ x (a > 1) is [M, n], then the value range of a is
Known: if the definition field and value field of function f (x) = a ^ x (a > 1) are [M, n]. Find the value range of A. please answer carefully.)
F (x) = a ^ x is an increasing function of X, that is, a ^ m = m, a ^ n = n, that is, a ^ x = x has two unequal roots, let g (x) = a ^ x-x, the original problem is equivalent to G (x) = 0 has at least two unequal roots, G '(x) = LNA * a ^ X-1, let g' (x) > 0, x > log (a) (1 / LNA); let g '(x)
The sum of the first n terms of an is SN. It is known that any n belongs to a positive integer, and the points (n, Sn) are all on the image of the function y = 3 * 2 ^ x + R
(1) Find R
(2) Let BN = 3N / an, (n is a positive integer), find the first n terms and TN of the sequence
Let {an} be Q
X = 1, y = S1 = A1, x = 2, y = S2 = a1 + A2, x = 3, y = S3 = a1 + A2 + a3, respectively
a1=6+r (1)
a1+a2=12+r (2)
a1+a2+a3=24+r (3)
(2)-(1)
a2=6
(3)-(1)
a2+a3=18
a3=18-a2=18-6=12
q=a3/a2=12/6=2
a1=a2/q=6/2=3
r=a1-6=3-6=-3
The general formula of sequence {an} is an = 3 × 2 ^ (n-1) = (3 / 2) 2 ^ n
bn=3n/an=2n/2^n
Tn=2(1/2^1+2/2^2+3/2^3+...+n/2^n)
Tn/2=2[1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)]
Tn-Tn/2=Tn/2=2(1/2^1+1/2^2+...+1/2^n-n/2^(n+1)]
Tn=4[1/2^1+1/2^2+...+1/2^n-n/2^(n+1)]
=4[(1/2)(1-1/2^n)/(1-1/2)]-2n/2^n
=4-4/2^n-2n/2^n
=4-(2n+4)/2^n
=4-(n+2)/2^(n-1)
The sum of the first n terms of an is SN. It is known that any n belongs to a positive integer, and the point (n, Sn) is on the image of the function y = 3 * 2 ^ x + R, then Sn = 3 * 2 ^ n + R
S1=a1=6+r
For n > = 2, s (n-1) = 3 * 2 ^ (n-1) + R
an=Sn-S(n-1)=3*2^(n-1)
a2=3*2=6
a3=3*2^2=12
Because it's an equal ratio sequence, q = 2
Then A2 / A1 = 2 = 6 / (... Expansion)
The sum of the first n terms of an is SN. It is known that any n belongs to a positive integer, and the point (n, Sn) is on the image of the function y = 3 * 2 ^ x + R, then Sn = 3 * 2 ^ n + R
S1=a1=6+r
For n > = 2, s (n-1) = 3 * 2 ^ (n-1) + R
an=Sn-S(n-1)=3*2^(n-1)
a2=3*2=6
a3=3*2^2=12
Because it's an equal ratio sequence, q = 2
So A2 / A1 = 2 = 6 / (6 + R)
So r = - 3
2.an=3*2^(n-1)
Bn=3n/an=n/(2^(n-1))
Tn=1/1+2/2+3/4+.....+n/(2^(n-1))①
Tn/2=1/2+2/(2*2)+3/(8)+...+n/2^n②
① (2) by deleting the same item, the
Tn/2=1/1+1/2+1/4+1/8+...+1/(2^(n-1))-n/2^n;
TN / 2 = (1-1 / 2 ^ n) * 2-N / 2 ^ n
Tn=(1-1/2^n)*4-n/2^(n-1)
n=1,T1=B1=2-1=1
Put it away
Let f (x) be an odd function with period 3 over R. if f (1) < 1, f (2) = 2a-1 / A + 1, then
A、a
It's option C
If f (x) is an odd function with a period of 3 and a domain of R, then f (x) is an odd function with a period of 3
f(x+3)=f(x),f(-x)=-f(x).
f(2)=f[3+(-1)]=f(-1)=-f(1).
If f (1) - 1,
F (2) = - f (1) > - 1, that is,
(2a-1)/(a+1)>-1,
(2a-1+a+1)/(a+1)>0,
3a/(a+1)>0.
a> 0 or a
C
f(2)=f(-1)=-f(1)>-1
Solving inequality
The sum of the first n terms of positive term sequence {an} is SN. For any n belonging to positive integer, the point (an, Sn) is on the image of function f (x) = x2 / 4 + X / 2
The sum of the first n terms of the positive term sequence {an} is SN. For any n belonging to a positive integer, the point (an, Sn) is on the image of the function f (x) = x2 / 4 + X / 2?
Sn=an^2/4+an/2
a1=s1=a1^2/4+a1/2---?4=a1+2--> a1=2
n>1,an=Sn-S(n-1)=1/4[an^2-a(n-1)^2]+1/2[an-a(n-1)]
2 = an-a (n-1)
The first term of {an} is 2 and the tolerance is 2
So an = 2n
Later, write a difference between two expressions, factorize, and you will get (an + an-1) (an + an-1-2) = 0, and you can get a tolerance of 2, and then let n = 1 to get the first term. Do it, and you'll find the effect of this condition - positive sequence
Don't ask me for the answer.....
By substituting sn-sn-1 = an, we can get the solution
Sn=an^2/4+an/2
n> When Sn = 2, sn-1 = a (n-1) ^ 2 / 4 + a (n-1) / 2
By subtracting the two formulas and factoring them, we can get an = a (n-1) + 2
An is the arithmetic sequence, and A1 = 2
an=2n
What does x2 / 4 mean?
Sn = (an) ^ 2 / 4 + an / 2, using the difference s (n + 1) - Sn, we can get the relationship between an and an + 1: an + 1-an = 2, when n = 1, S1 = A1, bring in the function expression, and get A1 = 2, so the sequence is a sequence with 2 as the first term and 2 as the tolerance, so the general term formula is an = 2n
2n
Given that the range of function f (x) = - 2acos (2x - π / 3) + 2A + B in the interval [0, π / 2] is [- 5,1], the values of constants A and B can be obtained
Given that the range of function f (x) = - 2acos (2x - π / 3) + 2A + B in the interval [0, π / 2] is [- 5, 1], the values of constants A and B are obtained. (can be ignored)
② Let g (x) = f (x + π / 2) and LGG (x) > 0, find the monotone interval of G (x).
When x = π / 6, the function f (x) has a minimum
f(x)=-2a*1+2a+b=b=-5
When x = π / 2, the function f (x) has a maximum
f(x)=-2a*（-1/2）+2a+b=3a+b=1
Substituting B = - 5 into the above formula, a = 2 is obtained
Conclusion a = 2, B = - 5
A = 2 b = - 5 question: the main question is the second question