It is known that the sum of the first n terms of the sequence {an} is Sn = N2 (n ∈ n *), the sequence {BN} is an equal ratio sequence, and satisfies B1 = A1, 2b3 = B4 (1) the general term formula of the sequence {an}, {BN}; (2) the sum of the first n terms of the sequence {anbn}

It is known that the sum of the first n terms of the sequence {an} is Sn = N2 (n ∈ n *), the sequence {BN} is an equal ratio sequence, and satisfies B1 = A1, 2b3 = B4 (1) the general term formula of the sequence {an}, {BN}; (2) the sum of the first n terms of the sequence {anbn}

(1) When n ≥ 2, an = sn-sn-1 = N2 - (n-1) 2 = 2N-1, so an = 2N-1 (n ∈ n *) let B1 = A1 = 1, let the common ratio of {BN} be q, let 2b3 = B4, then 2q2 = Q3, so q = 2, so BN = 2N-1 (2) let the sum of the first n terms of {anbn} be TN, then TN = 1 ×
Given that the standard deviation of a set of data x1, X2, X3, x4.x5 is 4, what is the variance of data (x1-4), (x2-4), (x3-4), (x4-4), (x5-4)?
Sixteen
Given that the function f (x) = x3-ax-1 increases monotonically on (- ∞, + ∞), the value range of real number a is obtained
∵ f (x) = x3-ax-1, ∵ f ′ (x) = 3x2-a, ∵ function f (x) = x3-ax-1 monotonically increases on (- ∞, + ∞), and ∵ f ′ (x) = 3x2-a ≥ 0 is constant, that is, △ ≤ 0, ∵ 12a ≤ 0, the solution is a ≤ 0, so when f (x) monotonically increases on (- ∞, + ∞), the range of a is (- ∞, 0]
What angle is the sine value of 0.4117
24 degrees, 18 minutes, 42 seconds
The sequence {an} and {BN} are known. If the two adjacent terms an and an + 1 of the sequence {an} are two of the equations x ^ 2-2 ^ NX + BN = 0 (n ∈ n *), and A1 = 1 (1)
(1) Finding the formula of sequence {an} and {BN}
(2) Let Sn be the sum of the first n terms of the sequence {an}, and ask whether there is a constant λ, such that BN xsn > 0
For any n ∈ n *, the value range of λ can be obtained if it exists; if it does not exist, please explain the reason
(1) When n is even, an = (2 ^ n-1) / 3, the algorithm is as follows: (a1 + A2) - (A2 + a3) + +(an-1+an)=2-2^2+2^3-2^4+…… +2 ^ (n-1) - 2 ^ n; when n is odd, an = (2 ^ n + 1) / 3 is the same; when n is even, BN = (2 ^ n-1) * (2 ^ (n + 1) + 1) / 9; when n is odd, BN = (2 ^ n + 1) * (2 ^ (n + 1) - 1) / 9
(2) There is no λ value
Calculation formula of variance and standard deviation~
The standard deviation is the result of square root of variance
Suppose the average value of this set of data is m
Variance formula s ^ 2 = 1 / n [(x1-m) ^ 2 + (x2-m) ^ 2 +... + (xn-m) ^ 2]
As shown in the picture!
Let f (x) = (x + 1) ln (x + 1). If f (x) ≥ ax holds for all x ≥ 0, find the value range of real number a
Solution 1: Let G (x) = (x + 1) ln (x + 1) - ax, find the derivative of the function g (x): G '(x) = ln (x + 1) + 1-A, let g' (x) = 0, the solution is x = ea-1-1, (I) when a ≤ 1, for all x > 0, G '(x) > 0, so g (x) is an increasing function on [0, + ∞), and G (0) = 0, so for X ≥ 0, there is g (x) ≥ g (0), that is, when a ≤ 1, for all x ≥ 0, there is g (0) F (x) ≥ ax. (II) when a ＞ 1, for 0 ＜ x ＜ ea-1-1, G ′ (x) ＜ 0, so g (x) is a decreasing function in (0, ea-1-1), and G (0) = 0, so for 0 ＜ x ＜ ea-1-1, G (x) ＜ g (0), that is, when a ＞ 1, not for all x ≥ 0, f (x) ≥ ax holds. In conclusion, the value range of a is (- ∞, 1]. Solution 2: Let G (x) = (x + 1) LN（ If the inequality f (x) ≥ ax holds, then G (x) ≥ g (0) holds. For G (x), find the derivative: G ′ (x) = ln (x + 1) + 1-A, let g ′ (x) = 0, get x = ea-1-1, when x > ea-1-1, G ′ (x) > 0, G (x) is an increasing function, when - 1 < x < ea-1-1, G ′ (x) < 0, G (x) is a decreasing function, so g (x) ≥ g (0) is necessary and sufficient for all x ≥ 0 If ea-1-1 ≤ 0, then a ≤ 1, that is, the value range of a is (- ∞, 1]
Who knows the sine of 0.5 degree?
Who knows what the sine of 0.5 degrees is?
Is there a unit for this? If so, what is it? Mm? Cm? Or
Sin (0.5 degrees) = 0.479425539
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Press the computer to pull OK!
In {an}, A1 = 1 / 2, an + 1 = Nan / (n + 1) (Nan + 1), n ∈ positive integer, let BN = 1 / Nan, prove that (1) {BN} is the expression of arithmetic sequence (2) Sn
(1)a(n+1)=nan/(n+1)(nan+1),
Transfer, (n + 1) a (n + 1) = Nan / (Nan + 1)
1 / (n + 1) a (n + 1) = 1 + 1 / Nan
BN = 1 / Nan, so B (n + 1) - BN = 1, B1 = 1 / (1 / 2) = 2
That is BN = 1 + N, is the arithmetic sequence
(2)
an=1/n(n+1)=1/n -1/(n+1)
Sn=1-1/2+1/2-1/3…… -1/(n+1)
=1-1/(n+1)
(1) Because an + 1 = Nan / (n + 1) (Nan + 1), 1 / an + 1 = n + 1 + (n + 1) / Nan
So d = BN + 1-bn = 1 / (n + 1) an + 1-1 / Nan = [n + 1 + (n + 1) / Nan] / (n + 1) - 1 / Nan = 1 + 1 / nan-1 / Nan = 1
And B1 = 1 / A1 = 2, so {BN} is an arithmetic sequence with B1 = 2 as the first term and d = 1 as the tolerance
(2) Is Sn the sum of sequence an or BN? Here we calculate b... expansion
(1) Because an + 1 = Nan / (n + 1) (Nan + 1), 1 / an + 1 = n + 1 + (n + 1) / Nan
So d = BN + 1-bn = 1 / (n + 1) an + 1-1 / Nan = [n + 1 + (n + 1) / Nan] / (n + 1) - 1 / Nan = 1 + 1 / nan-1 / Nan = 1
And B1 = 1 / A1 = 2, so {BN} is an arithmetic sequence with B1 = 2 as the first term and d = 1 as the tolerance
(2) Is Sn the sum of series an or BN? Let's calculate the sum of BN here,
From (1), we know that BN is an arithmetic sequence, Sn = NB1 + n (n-1) d / 2 = 2n + n (n-1) / 2 = (n * n + 3n) / 2
Given that the standard deviation of data x1, X2, X3, x4, X5 is 4 and the average is x, what is the sum of squares of the differences between the data and X
Junlang lieying team for you
S^2=16
S ^ 2 = 1 / 5 〔 (x1-x) ^ 2 + +(x5-x) ^ 2]
The sum of the squares of the differences between the data and X is 5 × 16 = 80