If the sum of the arithmetic sequence is 2380, the tolerance is 1 and the tolerance is 3, then there are several items in the sequence

If the sum of the arithmetic sequence is 2380, the tolerance is 1 and the tolerance is 3, then there are several items in the sequence

The first term is 1, the tolerance is 3, the nth term is 3n-2
The sum of the first n terms = (1 + 3n-2) n / 2 = (3n-1) n / 2
Let (3n-1) n / 2 = 2380
It is reduced to 3N & # 178; - N - 4760 = 0
Calculation by root formula

We know that the sequence {an (n subscript)} satisfies A1 (1 subscript) = 1, A2 (2 subscript) = 3, and prove that BN (n subscript) is an arithmetic sequence
Given sequence {an (n subscript)} satisfies A1 (1 subscript) = 1, A2 (2 subscript) = 3, an + 2 (n + 2 subscript) = 3an + 1 (n + 1 subscript) - 2An (n subscript) (n ∈ n *)
If the sequence {BN (n subscript)} satisfies: 4 ^ B1 (1 subscript) - 1 * 4 ^ B2 (2 subscript) - 1 *. 4N ^ B-1 = {an + 1} (n + 1 subscript) ^ BN, (n subscript) (n ∈ n *), it is proved that BN (n subscript) is an arithmetic sequence
(^ for superscript)

Analysis: B (n) must first ask for a (n); a (n) is easy to find! But there should be a problem with your second formula. Referring to your original formula, I found that B & #; actually solved a logarithm, which is very rare in high school mathematics! Note: the exam is not to defeat anyone, so when you find a very unsatisfactory number, you should think that there may be a problem, I think your second formula should be 4 ^ [b (1) - 1] * 4 ^ [b (2) - 1] *... * 4 ^ [b (n) - 1]) - n] = [a (n) + 1];
a(n+2)=3a(n+1)-2a(n)
That is, a (n + 2) - A (n + 1) = 2 [a (n + 1) - A (n)],
[a(n+2)-a(n+1)]/[a(n+1)-a(n)]=2;
Let C (n) = a (n + 1) - A (n); C &; = A &; - A &; = 3-1 = 2;
Obviously, C (n) is an equal ratio sequence with 2 as the first term and 2 as the common ratio
Then C (n) = 2 * (2 ^ (n-1)) = 2 ^ n
So a (n + 1) - A (n) = 2 ^ n;
Then a (n) - A (n-1) = 2 ^ (n-1) First item
a(n-1)-a(n-2)=2^(n-2)……………… Second
a(n-2)-a(n-3)=2^(n-3)……………… Third
……
……
……
……
a₂-a₁=2 …………………… Item n-1
In the equation, the left side is added with the left side, and the right side is added with the right side
a(n)-a₁=2^(n-1)+2^(n-2)+………… +2
That is, a (n) - 1 = - 2 (1-2 ^ (n-1))
a(n)=2^n-1;
Let's find B (n);
4^(b₁-1)*4^(b₂-1)*………… *4^[b(n)-1]=[a(n)+1]^b(n)
Then 4 ^ [B &; + B &; + +B (n) - n] = 2 ^ [n * B (n)]; let n = 1 give B & # 8321; = 2;
There are 2 [B &; + B &; +...] +b(n)-n]=n*b(n)………… In duplicate
2[b₁+b₂+………… +b(n-1)-(n-1)]=2^[(n-1)*b(n-1)]………… Two forms
By subtracting two from one, we can get the following result:
2[b(n)-1]=n*b(n)-(n-1)*b(n-1)
(n-1) * B (n-1) = (n-2) * B (n) + 2; here b > = 2;
Divide both sides of the equation by (n-1) * (n-2) to get:
b(n-1)/(n-2)=b(n)/(n-1)+2/(n-2)-2/(n-1);
b(n-1)/(n-2)-b(n)/(n-1)=2/(n-2)-2/(n-1);………… First item
b(n-2)/（n-3)-b(n-1)/(n-2)=2/(n-3)-2/(n-2);………… Second
…………
…………
…………
…………
b₂/1-b₃/2=2/1-1;……………… Item (n-2); in this case, n takes 3
The same equation is obtained by adding left and right and eliminating them;
b(n)=(n-1)*b₂+2-2*(n-1)=n*b₂-b₂-2*n+4;
Then B (n + 1) = n * B &; - 2 * n + 2;
Using B (n + 1) - B (n) = B &; - 2;
No matter what value B &; takes, (B &; - 2) is a constant!
A key university! Happy!

For indefinite integral: ∫ (Sin & # 178; 7x / (TaNx + cosx) DX,

No

Find the range of function y = √ 3sinx + cosx + 3

Y = √ 3sinx + cosx + 3 = 2Sin (x + π / 6) + 3 [Auxiliary Angle Formula]
∵sin(x＋π/6)∈[－1,1]
∴2sin(x＋π/6)∈[－2,2]
So y = 2Sin (x + π / 6) + 3 ∈ [1,5]
That is to say, the range of function y = √ 3sinx + cosx + 3 is [1,5]

If x is the smallest internal angle in a triangle, then the range of the function y = 3sinx + cosx is ()
A. (3，2]B. （1，2]C. (2，2]D. （1，2）

Y = 3sinx + cosx = 2 (32sinx + 12cosx) = 2Sin (x + π 6), ∵ x is the minimum internal angle of the triangle, ∵ 0 ＜ x ＜ π 3, ∵ π 6 ＜ x + π 6 ＜ π 2, ∵ 12 ＜ sin (x + π 6) ≤ 1, i.e. 1 ≤ 2Sin (x + π 6) ≤ 2, then the value range of the function is (1,2]

The range of the function y - √ 3sinx + cosx in the interval [- Π / 6, Π / 6} is?

y=√3sinx+cosx
=2sin(x+∏/6)
Let x + Π / 6 = t, then t belongs to [0, Π / 3]
Sint belongs to [0, √ 3 / 2]
The range of Y on [- Π / 6, Π / 6} is [0, √ 3]

The range of the function y = 3sinx-cos2x is

The range of the function y = 3sinx-cos2x is
Analysis: ∵ f (x) = 3sinx-cos2x
Let f '(x) = 3cosx + 2sin2x = cosx (3 + 4sinx) = 0
Cosx=0==>x1=2kπ-π/2,x2=2kπ+π/2
Sinx=-3/4==>x3=2kπ-arcsin(3/4),x4=(2k+1) π+arcsin(3/4)
f”(x)=-3sinx+4cos2x==>f”(x1)0
The maximum value of F (x) is taken at, x1, X2, f (x1) = - 2, f (x2) = 4
The minimum value of F (x) is taken at, X3, x4, f (x3) = f (x4) = - 9 / 4-1 + 2 * 9 / 16 = - 17 / 8
The range of the function y = 3sinx-cos2x is [- 17 / 8,4]

Finding the range of function y = 3sinx + cos2x-4
..

y=3sinx+cos2x-4
=3sinx+(1-2sin^2 x)-4
=-2sin^2 x + 3sinx -3
=-2(sinx-3/4)^2 +(9/16-3)
=-2(sinx-3/4)^2 - 39/16.
-If 1 ≤ SiNx ≤ 1, the maximum value of Y is 0 - 39 / 16 = - 39 / 16
If the function y (SiNx) = - 2 (sinx-3 / 4) ^ 2 - 39 / 16 of SiNx is a quadratic function with vertex on, then the minimum value can be obtained at one of the endpoint values - 1 and 1 of SiNx
When SiNx = - 1, y = - 2-3-3 = - 8;
When SiNx = 1, y = - 2 + 3-3 = - 2;
-8

Simple operation 666.66 * 6666.7 + 99999 * 22.222

666.66×6666.7+99999×22.222
=666.66×6666.7-22.222+100000*22.222
=30*22.222*6666.7-22.222+100000*22.222
=30*6666.7*22.222-22.222+100000*22.222
=(200001-1)*22.222+100000*22.222
=300000*22.222
=6666600

How to describe their rules in words

The first number is a six
The second number is two sixes and sixty-six
The third number is 666
.
The nth number is n 666.6
；；；；；；
N