How to find the sum of equal ratio sequence, (1 / 2) ^ (2n-1) and the final result is (2 / 3) [1 - (1 / 4) ^ n]?

How to find the sum of equal ratio sequence, (1 / 2) ^ (2n-1) and the final result is (2 / 3) [1 - (1 / 4) ^ n]?

When the first term is 1 / 2 (n = 1), the common ratio is 1 / 4. The result of the sum of the first n terms is [1 / 2 (1 - (1 / 4) ^ n)] / (1-1 / 4) = (2 / 3) [1 - (1 / 4) ^ n]
From the stem of the question, we know that A1 = 1 / 2, q = 1 / 4, Sn = A1 (1-Q ^ n) / (1-Q) can be obtained by substituting it into the formula
1 / 2 [1 - (1 / 4) ^ n] / (1-1 / 4) calculation reduction (2 / 3) [1 - (1 / 4) ^ n]
The common ratio of (1 / 2) ^ (2n-1) is (1 / 2) ＾ 2
Substitution formula Sn = A1 * (1-Q ^ n) / (1-Q)
Sn=1/2*(1-(1/4)^n)/(1-1/4)
Sn=(2/3)[1-(1/4)^n]
The common ratio of (1 / 2) ^ (2n-1) is (1 / 2) ＾ 2
Using the sum formula of equal ratio sequence to get the sum
1/2+（1/2）＾3+。。。 +(1/2)^(2n-1)
＝[1/2-(1/2)^(2n-1)*（1/2）＾2]/[1-(1/2)^2]
=(2/3)[1-(1/4)^n]
Because the common ratio of the original sequence is 1 / 4, Sn = (1 / 2 - ((1 / 2)) ^ (2n-1) × 1 / 4) / (1-1 / 4) = (2 / 3) [1 - (1 / 4) ^ n]
It is known that {an} is an arithmetic sequence with non-zero tolerance, and {BN} is an isobaric sequence, where a1 = 3, B1 = 1, A2 = B2, 3A5 = B3. If there exists a constant u, V and an = 3logubn + V for any positive integer n, then u + v=______ .
The tolerance of {an} is D, and the common ratio of {BN} is Q, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\whenn = 1, 3-V = logu1 = 0, ｜ v = 3 When n = 2, 12-3-3 = logu93, U6 = 93, u = 3, U + V = 6
It is known that log (1 / 7) [log (3) (log (2) x)] = 0
It is known that Log1 / 7 [log3 (log2x)] = 0 (1 / 7; 3; 2 are all base numbers), x ^ (- 1 / 2)=
log1/7[log3(log2x)]=0=log1/7(1)
So log3 (log2x) = 1
log3(log2x)=log3(3)
log2(x)=3
x=2³
X=8
Solve the equation first, layer by layer, from the outside to the inside
Outermost layer log (1 / 7) [log (3) (log (2) x)] = 0
We can get log (3) (log (2) x) = (1 / 7) ^ 0 = 1
Then we get log (2) x = 3 ^ 1 = 3
Finally, x = 2 ^ 3 = 8
So x ^ (- 1 / 2) = 8 ^ (- 1 / 2) = 1 / (2 √ 2) = (√ 2) / 4
log1/7[log3(log2x)]=0
Then iog1 / 7 (1) = o
So iog3 (iog2x) = 1 (iog2x = 3, iog3 (3) = 1)
Solid x = 8
^ - 2) = (- x)
When the line L passes through point a (1,2) and the area of the triangle surrounded by the positive half axis of the two coordinate axes is 4, the equation of the line L is obtained
Let X / A + Y / b = 1 (where a > 0 and b > 0),
Because the straight line passes (1,2), then 1 / A + 2 / b = 1,
If the triangle area is 4, then AB / 2 = 4, that is, ab = 8,
A = 2, B = 4,
So the linear equation is x / 2 + Y / 4 = 1, that is, 2x + y-4 = 0
Let Y / L = a + 1
Line L passes through point a (1,2) - > 1 / A + 2 / b = 1 - > b + 2A = ab
The area of the triangle surrounded by the positive half axis of the two coordinate axes is 4 → a > 0, b > 0 → AB / 2 = 4 → AB = 8 → B = 8 / A
Bring in B + 2A = ab → 8 / A + 2A = 8 → 2A ^ 2-8a + 8 = 0 → a = 2 → B = 4
So the line L: X / A + Y / b = 1 → X / 2 + Y / 4 = 1 → 2x + y-4 = 0
y=-2x+4
Log base 5 times (log base 4 (log base 3 (x))) = 0
Log base 5 times (log base 4 (log base 3 (x))) = 0
Base 4 of log (base 3 (x)) = 1
The base of log is 3 (x) = 4
The fourth power of x = 3
x=81
analysis
log5(log4(log3x))=0
therefore
log4 (log3x)=1
therefore
log3x=4
x=3^4
=81
Base 4 of log (base 3 (x)) = 5 ^ 0 = 1
Log base 3 (x) = 4 ^ 1 = 4
x=3^4=81
Given that the line L passes through P (1,2), and the area of the triangle enclosed by the positive half axis of the two coordinate axes is the smallest, find L
2. Given that the straight line m passes through the point Q (5,3), and the intercept on the coordinate axis is equal, the equation of M is obtained
Let the equation of a straight line be Y-2 = K (x-1), then it intersects with two coordinate axes
(0,2-k),(1-2/k,0)
Area = 1 / 2 * (2-k) (1-2 / k) = 1 / 2 * (- K-4 / K + 4)
If and only if k = 4 / K (k
1 l:y=2x
2 m:y1=-x+2
or y2=x-2
hjnjh j
Log base 2 (log base 3 (log base 4 x)) = log base 3 (log base 4 (log base 2 y)) = log base 4
(log base 2 (log base 3 z)) = 0, then x + y + Z equals?
I've written a lot about this. I hope that those who do will not find it troublesome
log2（log3(log4x)=0
（log3(log4x)=2^0=1
log4x=3^1=3
x=4^3=64
log3（log4(log2y)=0
log4(log2y)=3^0=1
log2y=4^1=4
y=2^4=16
log4（log2(log3z)=0
log2(log3z)=4^0=1
log3z=2^1=2
z=3^2=9
So x + Y-Z = 64 + 16-9 = 71
According to the original play will be better search
If the intercept of the first-order function y = KX + B on the y-axis is - 4, and the area of the triangle bounded by the coordinate axis is 4, then the analytic expression of the first-order function is
The y-intercept is - 4, that is, B = - 4, so the image of the function intersects with the y-axis at point B (0, - 4), and the image intersects with the x-axis at point a (4 / K, 0). The area of the triangle formed by the image and the coordinate axis is 4. So s ⊿ OAB = 1 / 2 | OA | ob | = 1 / 2 | 4 / K | * | - 4 | = 4, so k = 2. So the analytic formula of this first-order function is y = 2X-4
y=2x-4
y=-2x-4
Given that a and B belong to (0,1), M = | log B (1-A) |, n = | log B (1 + A + A ^ 2 + A ^ 3 +. + A ^ 2009) |, then the relationship between M and N is
Urgent solution!
a. B belongs to (0,1), 1 + A + A ^ 2 + A ^ 3 +. + A ^ 2009 = (1-A ^ 3000) / (1-A) 0, y = | log B (x) | is a decreasing function, (1-A ^ 3000) / (1-A) > (1-A), so m > n
m=|log b(1-a)|=|logb+log(1-a)|
m^2=(logb)^2+(log(1-a))^2+2*logb*log(1-a)
n=|log b(1+a+a^2+a^3+.......+a^2009)|=|logb+log[(1-a^2010)/(1-a)]|=
=|logb+log(1-a^2010)-log(1-a)|
N ^ 2 = (logb) ^ 2... Expand
m=|log b(1-a)|=|logb+log(1-a)|
m^2=(logb)^2+(log(1-a))^2+2*logb*log(1-a)
n=|log b(1+a+a^2+a^3+.......+a^2009)|=|logb+log[(1-a^2010)/(1-a)]|=
=|logb+log(1-a^2010)-log(1-a)|
n^2=(logb)^2+[log(1-a^2010)]^2+[log(1-a)]^2-2*logb*log(1-a)+2*logb*log(1-a^2010)-2*log(1-a^2010)*log(1-a)
n^2-m^2=-4*logb*log(1-a)+[log(1-a^2010)]^2+2*logb*log(1-a^2010)-2*log(1-a^2010)*log(1-a)
n^2-m^2=2logb*[log(1-a^2010)-2log(1-a)]+2log(1-a^2010)(log(1-a^2010)-log(1-a))
a^20101-a^2010>1-a
Log (1-A ^ 2010) > log (1-A) but
|log(1-a^2010)|
In the circuit as shown in the figure, the power supply voltage remains unchanged. When the switches S1 and S3 are closed and S2 is opened, the current representation is 1.5A;
When S1 is closed and S2 and S3 is disconnected, the current indication is 0.5A; when S2 is closed and S1 and S3 is disconnected, the power consumed by R2 is 2W. Find: (1) the resistance value of resistance R2. (2) what is the power supply voltage. (3) in the figure, if R2 is replaced with "50 Ω 2.5A" sliding rheostat, the ammeter range is "0-3A". When S1 is closed and S3 is disconnected, what is the minimum value of sliding rheostat connected into the circuit?
-----------Resistance R2 -------- switch S1---
|Resistance R1|
| | |
---------------Switch S3 ------|
Ammeter switch S2|
------------------Power supply-------------------------------
Can't see where the S2 switch is
(1) When S1 and S3 are closed and S2 is opened, R1 and R2 are connected in parallel, I = 1.5A; when S1 is closed and S2 and S3 are opened, there is only R1 in the circuit, I = 0.5A
After two comparisons, the power supply voltage U remains unchanged, and there is no influence on R1 before and after parallel connection, I1 = 0.5A
∴I2=I-I1=1.5A-0.5A=1A
∵ R1, R2 in parallel
∴I1：I2=R2：R1=1：2
Ψ R1 = 2r2, u = i1r1 = 0.5A × R1 -------... Expansion
(1) When S1 and S3 are closed and S2 is opened, R1 and R2 are connected in parallel, I = 1.5A; when S1 is closed and S2 and S3 are opened, there is only R1 in the circuit, I = 0.5A
After two comparisons, the power supply voltage U remains unchanged, and there is no influence on R1 before and after parallel connection, I1 = 0.5A
∴I2=I-I1=1.5A-0.5A=1A
∵ R1, R2 in parallel
∴I1：I2=R2：R1=1：2
∴R1=2R2，U=I1R1=0.5A×R1-------①
When S2 is closed and S1 and S3 are disconnected, R1 and R2 are connected in series, P2 = 2W
U=I（R1+R2）=P2U2×(R1+12R1 )---------②
The results are as follows
0.5A×R1=P2U2×(R1+12R1 )=P2U2×(R1+12R1 )=2WU2×32R1
The solution is: U2 = 6V,
∵ R1 and R2 are connected in series, R1 = 2r2, ∵ U1 = 12V, u = 18V
U=I1R1=0.5A×R1=18V，R1=36Ω，R2=18Ω
(2) If the sliding rheostat is replaced, when S1 and S3 are closed and S2 is disconnected, the sliding rheostat and R1 are connected in parallel. It is known that I1 = 0.5A, u = 9V
When I = I1 + I2 = 3A, I2 = 2.5A
Ψ 9vr small = 2.5A
Ψ r small = 3.6 Ω
Answer: (1) the resistance value of resistance R2 is 18 Ω;
(2) The power supply voltage is 18V;
(3) The minimum resistance of sliding rheostat connected into the circuit is 3.6 Ω
I forgot the formula
I can't write formulas
You first calculate the ratio of resistance R2 to R1, and then you can work it out. When S1 and S3 are closed and S2 is opened, the total resistance is one-third of that when S1 is closed and S2 and s3r1 is opened. Then the resistance ratio of R1 and R2 can be calculated.
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When S1 and S3 are closed and S2 and S3 are disconnected, the current indication is 1.5A; when S1 is closed and S2 and S3 are disconnected, the current indication is 0.5A; it indicates that resistance R2 = 2r1. The relationship between resistance and voltage can also be obtained.
When S2 is closed and S1 and S3 are disconnected, the power consumed by R2 is 2W. I * I * R2 is 2W. I is the current. I=U/（R1+R2）
In fact, there are three formulas: 1: U / r2 = 0.52: U / (R1 + R2) * U / (R1 + R2) * R2 = 23; U /
When S1 and S3 are closed and S2 and S3 are disconnected, the current indication is 1.5A; when S1 is closed and S2 and S3 are disconnected, the current indication is 0.5A; it indicates that resistance R2 = 2r1. The relationship between resistance and voltage can also be obtained.
When S2 is closed and S1 and S3 are disconnected, the power consumed by R2 is 2W. I * I * R2 is 2W. I is the current. I=U/（R1+R2）
In fact, there are three formulas: 1: U / R 2 = 0.52: U / (R 1 + R 2) * U / (R 1 + R 2) * r 2 = 23; U / (R 1 + R 2) = 1.5: R 1 = 18 Ω, R 2 = 9 Ω and u = 9V
Considering no over current, the maximum current of sliding rheostat can only be 3-U / R1 = 2, estimated U / 2 = 4.5, that is, the minimum current can not be less than 4.5 Ω