Note that the sum of the first n terms of the sequence {an} is Sn, and Sn = 2 (an-1), then A2=______ .

Note that the sum of the first n terms of the sequence {an} is Sn, and Sn = 2 (an-1), then A2=______ .

∵ Sn = 2 (an-1), ∵ S1 = 2 (A1-1), ∵ A1 = 2 ∵ S2 = 2 (A2-1) = 2 + A2 ∵ A2 = 4, so the answer is: 4
Given that α and β belong to (0, π / 2) and cos α = radical 5 / 5, cos β = radical 10 / 10, find α + β (important process)
Cos α = √ 5 / 5, cos β = √ 10 / 10 α, β belongs to (0, π / 2). From the square formula of trigonometric function with the same angle, we can get: sin α = 2 √ 5 / 5, sin β = 3 √ 10 / 10, so cos (α + β) = cos α * cos β - sin α * sin β = √ 5 / 5 * √ 10 / 10-2 √ 5 / 5 * 3 √ 10 / 10 = - √ 2 / 2, and α, β belong to (0, π / 2)
Let u = R, a = {X / y = log2 bottom (- x + 2x + 3)}, B = {Y / y = 2 x, (- infinite, 2)}, C = {X / a less than x, less than 2a-1} (1)
(1) Find a B
(2) If C is included in (a intersects b), find the value range of real number a
(1) . a = {X / y = log2 bottom (- x + 2x + 3)}
-X + 2x + 3 > 0
-1＜x＜3
A=（-1,3）
B = {x-square of Y / y = 2, (- infinity, 2)}
It can be drawn from the graph, 0 ＜ y ＜ 4
B=（0,4）
（2）A∩B=（0,3）
1, C is an empty set, then a < 1
2, C is not an empty set, then 1 < a < 2
In conclusion, the value range of a is (- infinity, 2)
Let the sum of the first n terms of the sequence {an} be Sn, if A1 = 1, an + 1 = 3Sn (n ∈ n *), then S6 = ()
A. 44B. 45C. 13•(46−1)D. 13•(45−1)
From an + 1 = 3Sn, an = 3sn-1 (n ≥ 2) is obtained. By subtracting the two formulas, an + 1-an = 3an, that is, an + 1 = 4An (n ≥ 2), A1 = 1, A2 = 3s1 = 3, a2a1 = 3,... A2, A3 The common ratio is 4, an = 1, n = 13.4n − 2, n ≥ 2, S6 = a1 + A2 + a3 + +a6=1+3+12+… +3.44 = 1 + 3 (1 − 45) 1 − 4 = 45, so B
Then π ∈ (β) / sin = 0,2, and α / sin = 2=
cosα=3/5 sinα=4/5
Sin (α + β) = - (radical 2) / 4
We know that α + β is in the fourth quadrant, cos (α + β) = - (radical 14) / 4
Turn the above two types into two types
Sin α COA β + cos α sin β = - (radical 2) / 4
Cos α COA β - sin α sin β = (radical 14) / 4
Namely:
4 / 5coa β + 3 / 5sin β = - (radical 2) / 4
3 / 5coa β - 4 / 5sin β = (radical 14) / 4
The solution is sin β = - 3 * (radical 2) / 20 - (radical 14) / 5
Let u = R, M = {y ∈ R | y = 2x, X ＞ 0}, n = {x ∈ R | 2x-x2 ＞ 0}, then m ∩ n is ()
M={y∈R|y=2^x，x＞0，N={x∈R|2x-x²＞0}
∵ function y = 2x, (x ＞ 0) has a range of Y ＞ 1,
The set M = {y ∈ R | y = 2x, X ＞ 0} = {y | y ＞ 1},
That is: all real numbers greater than 1 constitute a set M, which can also be written as M = {x | x ﹥ 1},
N = {x ∈ R | 2x-x2 ＞ 0} = {x ∈ R | x (X-2) ＜ 0} = {x | 0 ＜ x ＜ 2},
∩ m ∩ n = {x | 1 ＜ x ＜ 2}, expressed by interval as (1,2)
So choose a
Set M is the range of X of function y = 2 when x > 0, then:
M={y|y>1}
The set n is the solution set of inequality 2x - X & # 178; > 0, and we get: X & # 178; - 2x
The known sequence {an} satisfies A1 = 1 / 2, the first n terms and Sn = n ^ 2An
The known sequence {an} satisfies A1 = 1 / 2, the first n terms and Sn = n ^ 2An
1) Find A2, A3, A4
2) The general term formula of conjecture sequence {an} is proved by mathematical induction
It is known that the sequence {an} satisfies A1 = 1 / 2, the first n terms and Sn = n & sup2; an; (1) finding A2, A3, A4; (2) conjecturing the general term formula of the sequence {an}, which is proved by mathematical induction. (1) it is easy to get A2 = 1 / 6, A3 = 1 / 12, A4 = 1 / 20; (2) conjecturing an = 1 / [n (n + 1)]
a1=1/2
a2=1/6
a3=1/12
a4=1/20
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an=1/((n+1)*n)
As for proving that you've done it yourself, it's not difficult.
(1) A2 = s2-s1 = 4a2-1a1. Then A2 = 1 / 6
A3 = s3-s2 = 9a3-4a2. Then A3 = 1 / 12
A4 = s4-s3 = 16a4-9a3, then A4 = 1 / 20
(2) Conjecture: an = 1 / (n * (n * 1))
When n = 1, A1 = 1 / (1 * 2) = 1 / 2 holds
When n = k, K ≥ 2, AK = sk-s (k-1) = k ^ 2 * AK - (k-1) ^ 2 * a (k-1)
AK... Unfold
(1) A2 = s2-s1 = 4a2-1a1. Then A2 = 1 / 6
A3 = s3-s2 = 9a3-4a2. Then A3 = 1 / 12
A4 = s4-s3 = 16a4-9a3, then A4 = 1 / 20
(2) Conjecture: an = 1 / (n * (n * 1))
When n = 1, A1 = 1 / (1 * 2) = 1 / 2 holds
When n = k, K ≥ 2, AK = sk-s (k-1) = k ^ 2 * AK - (k-1) ^ 2 * a (k-1)
ak/a(k-1)=(k-1)/(k+1).
At the same time, AK / a (k-1) = 1 / (k * (K + 1) △ 1 / ((k-1) * k) = (k-1) / (K + 1), which holds
Therefore, it is proved
An = 1 / (n * (n * 1))
a2=1/6
a3=1/12
a4=1/20
an=1/[n(n+1)]
While doing Chinese class It should be true. If you prove it, just follow the steps of the book step by step: -)
If Sina = radical 10 / 10. B = arc cos (- radical 5 / 5). 0 < a < Π / 2, prove: a + B = 3 Π / 4, the gods help
We need to explain it in detail
The root sign is less than 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 10 / sincob = 5 / sincob = 10 / sincob = 10 / sincob = 5 / sincob = 10 / sincob / sincob = 10 / sincob = 5 / sincob / sincob = 10 / sincob = 5 / sincob / sincob / sin
Let a = {y | y = x & # 178; + 2x + 2, X ∈ r} and B = {y | (Y-2) (y + 3) ≤ 0}, then a ∩ B is equal to
A:y=x²+2x+2=(x+1)²+1≥1 y≥1
B:(y-2)(y+3)≤0 -3≤y≤2
A∩B={y|1≤y≤2},
Hello, the solution is as follows:
A = {y | y = x & # 178; + 2x + 2, X ∈ r}, a = {y | y ≥ 1, X ∈ r}
B = {y | (Y-2) (y + 3) ≤ 0}, B = {y | - 3 ≤ y ≤ 2}
So a ∩ B = {y | 1 ≤ y ≤ 2}
If you have any questions, please communicate. Follow up: OK, thank you!
Given that the sequence {an} satisfies an = 1 / 3Sn, Sn is the sum of the first n terms of an, and A1 = 1, find the general term formula of an
an=1/3Sn
a(n-1)=1/3s(n-1)=1/3(sn-an)
2 / 3an = a (n-1)
q=an/a(n-1)=3/2
N-1 power of an = A1 * q
That is, an is an equal ratio sequence
The general term is an = 3 / 2 ^ (n-1)