The sum of the first n terms of the sequence 1, - A, a ^ 2, - A ^ 3. Is equal to the sum of the proportional sequence (1 - (- a) ^ n) / (1 + a)? Why? Please write the process clearly

The sum of the first n terms of the sequence 1, - A, a ^ 2, - A ^ 3. Is equal to the sum of the proportional sequence (1 - (- a) ^ n) / (1 + a)? Why? Please write the process clearly

Let the sum of the first n terms be s, which is divided into three steps: when a = 0, it is not an equal ratio sequence, then s = 1, when a = - 1, s = n, when a is not 0 and - 1, s = 1-A + A ^ 2-A ^ 3 +... + (- a) ^ (n-1) ^
The common ratio is - A, and the first term is 1
It is known that O is the center of circumscribed circle of acute angle △ ABC, and ∠ a = θ, if (CoSb / Sina) * vector AB + (COSC / SINB) * vector AC = 2m * vector Ao,
Then M=
It seems that (CoSb / Sina) * vector AB should be (CoSb / sinc) * vector AB, which is more reasonable! If the radius of circumcircle is r, then: (CoSb / sinc) * vector AB + (COSC / SINB) * vector AC = 2m * vector Ao can be transformed into: (CoSb / sinc) * (vector ob vector OA) + (COSC / SINB) * (vector oc-oa) = - 2m * vector OA (*)
According to the principle of the number of the vector ad → + D →, we can simplify the vector ad → + D →, and calculate the vector ad → + D → In short, we use the sine theorem to transform and use trigonometric function to express m. We use the induction formula and the theorem of the sum of internal angles of triangles to get CoSb = - cos (a + C), which is substituted into the M formula, and then we use two... Expansion
Analysis: draw the corresponding figure according to the meaning of the question, take the midpoint of AB as D, according to the parallelogram rule of plane vector, we can get Ao → = ad → + do →, substitute it into the known equation, connect OD, we can get ad →⊥ ab →, we can get its scalar product as 0, multiply ab →onboth sides of the simplified equation, and then use the calculation method of vector module and the scalar product algorithm of plane vector to simplify the equation Then we use sine theorem to transform and express m with trigonometric function. Then we use induction formula and triangle's inner angle sum theorem to get CoSb = - cos (a + C), which is substituted into the M formula. Then we use cosine function formula of sum and difference of two angles to simplify and get the simplest result after offsetting and combining reduction. If we substitute ∠ a = θ, then we can express m with trigonometric function of θ →=AD→+DO→，
By substituting cosbsincab → + coscsinbac → = 2mao →
cosBsinCAB→+cosCsinBAC→=2m(AD→+DO→)，
From ad →⊥ ab →, do → &; ab → = 0,
Two sides multiply ab →
cosBsinCAB→•AB→+cosCsinBAC→•AB→=2m(AD→+DO→)•AB→=mAB→•AB→，
That is, cosbsincc2 + coscsinbbc &; cosa = MC2,
Starting from sine theorem Asina = bsinb = csinc
Given that the line L passes through the point a (- 4, - 2), and the point a is the midpoint of the line segment cut by the two coordinate axes, then the equation of the line L is --------···
Let: the coordinate of the intersection of the line and the x-axis be m (x, 0), and the intersection of the line and the y-axis be n (0, y). Because point a (- 4, - 2) is the midpoint of the line Mn, so - 4 × 2 = x + 0, x = - 8 - 2 × 2 = y + 0, y = - 4, so the line passes through (- 8,0) and (0, - 4). Let: the line is y = KX + B, so B = - 4, 0 = - 8K + B, k = - 0.5, so the line L is: y = - 0.5x-4
[calculation:] 3 times root [(26 / 27) - 1] - 3 times root -0.008; - | (- 3) & # 178; | - (1 / 3-1 / 4) & # 178; × √ (- 6) & # 178;
Root of 3 times [(26 / 27) - 1] - root of 3 times -0.008
=³√-1/27+³√0 .008
=-1/3+0.2
=-1/3+1/5
=-2/15
-|(-3)²|-(1/3-1/4)²×√(-6)²
=9-1/144x6
=9-1/24
=8 and 23 / 24
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If the area of the triangle formed by a straight line passing through a (- 2,2) and two coordinate axes is 1, the equation of the straight line is obtained
If the triangle area is 5, how to find it
Let the line be x / A + Y / b = 1, then its intersection with the coordinate axis is (0, b), (a, 0) so the triangle area = | ab | / 2 = 1 | ab | = 2, and the line crosses a (- 2,2) - 2 / A + 2 / b = 11 / B-1 / a = 1 / 2 (a-b) / (AB) = 1 / 2Ab = 2 (a-b) if a > b, then A-B > 0, then | ab | = AB = 2 = 2 (a-b) A-B = 1A = B + 1ab = 2B ^ 2 + B-2 = 0 (B + 2) (B-1)
Let the linear equation be y = KX + B, because the line passes through point a (- 2,2)
SO 2 = - 2K + B, B = 2 + 2K
y=kx+(2+2k)
When x = 0, y = 2 + 2K
When y = 0, KX + 2 + 2K = 0, x = - (2 + 2K) / K
The area of the triangle formed by the line and two coordinate axes is 1
So we have (1 / 2) | 2 + 2K | * | - (2 + 2K) / K | = 1
(2+2k)²=2|k|
When k > 0, 4... Expand
Let the linear equation be y = KX + B, because the line passes through point a (- 2,2)
SO 2 = - 2K + B, B = 2 + 2K
y=kx+(2+2k)
When x = 0, y = 2 + 2K
When y = 0, KX + 2 + 2K = 0, x = - (2 + 2K) / K
The area of the triangle formed by the line and two coordinate axes is 1
So we have (1 / 2) | 2 + 2K | * | - (2 + 2K) / K | = 1
(2+2k)²=2|k|
When k > 0,4 + 8K + K & sup2; = 2K, K & sup2; + 6K + 4 = 0, there is no solution
So K
Calculation: | 2-A | + | A-4 | + radical (A-2) & # 178; (2 < a < 3)
|2-A | + | A-4 | + radical (A-2) & # 178;
=|2-a|+|a-4|+|a-2|
Because 2
|2-A | + | A-4 | + radical (A-2) & # 178;
=a-2+4-a+a-2
=a
Do not understand can ask, help please adopt, thank you!
Solution: | 2-A | + | A-4 | + radical (A-2) & # 178;
=|2-a|+|a-4|+|a-2|
Because 2
Given that the line L passes through the point (1,2), the equation of L? (1) and the coordinate axis form the smallest triangle area in the first quadrant
There is a solution process. Urgent need!
So the equation Y-2 = K (x-1)
The intercept x0 on the x-axis = | (K-2) / K |, and the intercept Y0 on the y-axis = | (K-2) / K ||
When k is less than 0
x0*y0=k-2/k*(2-k)=|4-[k+4/k]|
So k = 4 / K, so k = - 2
L y-2=-2(x-1) 2x+y-4=0
Given the function f (x) = 3-2log2x, G (x) = log2x. (1) if x ∈ [1,4], find the range of function H (x) = (f (x) + 1) g (x); (2) find the maximum value of function m (x) = f (x) + G (x) − | f (x) − g (x) | 2; (3) if the inequality f (x2) f (x) > kg (x) in any x ∈ [1,4], the inequality holds, find the range of real number K
Let t = log2x (1) H (x) = (4-2log2x) · log2x = - 2 (t-1) 2 + 2, X ∈ [1,4], t ∈ [0,2]} H (x) be in the range of [0,2] (2) ∵ m (x) = g (x) & nbsp; & nbsp; f (x) ≥ g (x) f (x) & nbsp; & nbsp; f (x) < g (x) let the smaller value of F (x) and G (x) be m
The line passing through the fixed point (1.4) forms a triangle with the coordinate axis in the first quadrant, and the area is the smallest
Let the area enclosed by X / A + Y / b = 1 be in the first quadrant, so the intersection of the line and the coordinate axis is positive a > 0, b > 0 area = AB / 2. Substitute the fixed point into 1 / A + 4 / b = 1b + 4A = ABB + 4A = (B + 4a) * (1 / A + 4 / b), because 1 / A + 4 / b = 1 = B / A + 4 + 4 + 16A / b = 8 + (B / A + 16A / b) because a > 0, b > 0 is determined by the mean inequality B / A + 16A / b > = 2
The slope of a line must exist
Let the slope be K
So the linear equation is y-4 = K (x-1)
The intersection of the line and the x-axis is (1-4 / K, 0)
Intersection with Y-axis at (0,4-k)
Therefore, the triangle area s = (1-4 / k) (4-K) / 2 = (4-k-16 / K + 4) / 2 = 4 + (- K-16 / k) / 2
Because the triangle is in the first quadrant. So k = 8, when k = -... Expand
The slope of a straight line must exist
Let the slope be K
So the linear equation is y-4 = K (x-1)
The intersection of the line and the x-axis is (1-4 / K, 0)
Intersection with Y-axis at (0,4-k)
Therefore, the triangle area s = (1-4 / k) (4-K) / 2 = (4-k-16 / K + 4) / 2 = 4 + (- K-16 / k) / 2
Because the triangle is in the first quadrant, k = 8, when k = - 4, the maximum value is obtained
The triangle area is 8
So the linear equation is y-4 = - 4 (x-1)
In general, 4x + Y-8 = 0
The known function f (x) = log 0.5 (2-ax) / (x-1) (a is a constant, a
From (2-ax) / (x-1) > 0, (2-ax) * (x-1) > 0; if A0; and 2 / A1; if 0