If (M + 1) ^ 1 / 2

If (M + 1) ^ 1 / 2

1 / 2 is a positive number, so the power function is an increasing function
So 0

If (M + 1) ^ - 1

From (M + 1) ^ - 1-1,3-2m2 / 3, M > 2 / 3
2. If 3-2m3 / 2, M > 3 / 2
3. If M + 10 is constant, the solution is m3 / 2 or m

Let constant a ∈ R, set a = {x | (x-1) · (x-a) ≥ 0}, B = {x | x ≥ A-1}, if a ∪ B = R, then the value range of a is______ ．

When a ≥ 1, the solution of inequality in set a is: X ≤ 1 or X ≥ a, i.e. a = {x | x ≤ 1 or X ≥ a}, ∵ B = {x | x ≥ A-1}, and a ∪ B = R, ∪ A-1 ≤ 1, i.e. 1 ≤ a ≤ 2; when a < 1, the solution of inequality in set a is: X ≤ a or X ≥ 1, i.e. a = {x | x ≤ a or X ≥ 1}

Let constant a ∈ R, set a = {x | x (x-a) ≥ 0}, B = {x | x ≥ A-1}, if a ∪ B = R, then the value range of a is
The problem is really wrong. This is right
Let constant a ∈ R, set a = {x | (x-1) (x-a) ≥ 0}, B = {x | x ≥ A-1}, if a ∪ B = R, then the value range of a is

The title is wrong
Set a = {x | (x-a) ≥ 0} = {x | x ≥ a}
Set B = {x | x ≥ A-1}
Then set AUB = {x | x ≥ a} ≠ R
So the value range of a is an empty set

Given that A1 = 1 in the sequence {an}, and satisfied that an + an-1 is not equal to 0, Sn = 1 / 6 * (an + 1) (an + 2). (1) find the general term an, and explain what sequence {an} is
(2) Finding the first n terms and Sn of sequence {an}

Sn+1=1/6(a(n+1) +1)(a(n+1) +2)
Sn=1/6*(an+1)(an+2).
By subtracting the two formulas, we can get the following results
a(n+1)*a(n+1)-an*an=3*(a(n+1)+an)
The results are as follows
a(n+1)-an=3
So {an} is an arithmetic sequence with A1 = 1 as the first term and d = 3
So an = 1 + (n-1) * 3 = 3n-2

It is known that in the sequence {an}, A1 = - 1, the sum of the first n terms is Sn (not equal to 0), satisfying Sn * s (n-1) = an (n ≥ 2)

Guess, an = 1 / [n (n-1)] n > 1
When n = 2, A2 = 1 / 2
Suppose k = n, an = 1 / [n (n-1)]
When k = n + 1, Sn = 1 + 1 / 2 * 1 + +1/[n(n-1)]=2-1/n
an+1=(Sn+an+1)Sn
an+1=1/[n(n+1)]
Get proof

Given that an in {an} is not equal to 0 (n > = 1), A1 = 0.5, the first n terms and Sn satisfy: an = 2 (SN) ^ 2 / (2sn-1) (n > = 2), find the general term formula of {an}

2 (SN) ^ 2 / (2sn-1) = SN-S (n-1), 1 / sn-1 / S (n-1) = 2, that is {1 / Sn} is equal difference, n > 1, d = 2, 1 / S2 = 4
a2=S2-a1=2(S2)^2/2S2-1
S2=1/4
1/Sn=2n
an=1/2n(1-n)

In the equal ratio sequence {an}, A1 ＞ 1, and the first n terms and Sn satisfy LiMn →∞ Sn = 1A1, then the value range of A1 is ()
A. （1，+∞）B. （1，4）C. （1，2）D. （1，2）

From the meaning of the title, we know LiMn →∞ Sn = a11 − q = 1A1, ∵ A12 = 1-Q, ∵ A1 ＞ 1, | Q | 1, | 1 ＜ A12 ＜ 2, ∵ 1 ＜ A1 ＜ 2

In known sequence {an}, an = 1 / N (n + 2). How to find the limit of Sn when n tends to positive infinity?

an=1/n(n+2)=1/2*(1/n-1/(n+2))Sn=a1+a2+.+an=1/2*(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2))=1/2(1+1/2-1/(n+1)-1/(n+2))=1/2(3/2-(2n+3)/(n^2+3n+2)=(3n^2+5n)/2(n^2+3n+2)=(3+5...

In the sequence an, an = 1 / (n * (n + 1) * (n + 2)), then the limit of Sn is?
Shouldn't the last 1-2 + 1 / 2 be equal to - 1 / 4?

Firstly, we analyze an = 1 / (n * (n + 1) * (n + 2)) = 1 / 2 {1 / [n (n + 1)] - 1 / [(n + 1) (n + 2)]}
So Sn = a1 + A2 +... An
=1/(1*2*3)+1/(2*3*4)+...1/[n(n+1)(n+2)]
=1/2*[1/(1*2)-1/(2*3)]+1/2*[1/(2*3)-1/(3*4)]+..+1/2*[n(n+1)]-1/[(n+1)(n+2)]
=1/2{1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+n(n+1)-1/[(n+1)(n+2)]}
=1/2*[1/2-1/(n^2+3n+2)]
So limsn = Lim {1 / 2 * [1 / 2-1 / (n ^ 2 + 3N + 2)]} = LIM (1 / 4) = 1 / 4