1. 2, 4, 8, 16, 32, 64, 128... Who knows the formula

1. 2, 4, 8, 16, 32, 64, 128... Who knows the formula

The nth term is the N-1 power of 2, and N is a natural number greater than or equal to 1

What is the law of such numbers
A (1,3) A1 (2,3) A2 (4,3) A3 (8,3) A4 (16,3). An (?, 3) is expressed in an algebraic expression with N. don't just say the answer

They are the 0 th power of 2, the 1 th power of 2, the 2 th power of 2, the 3 th power of 2, the 4 th power of 2, and the 5 th power of 2

-2,4,-8,16,-32,64… What is the order of these numbers?

This is very simple. The n power of (- 1) is multiplied by the n power of 2

1, known a + 1 / a = 2, find a ^ 2 +! / A ^ 2, guess a ^ n + 1 / A ^ n (n is any natural number) is equal to? And prove your conclusion
2. Calculation: 1 / A-1 + 1 / (A-1) (A-2) + 1 / (A-2) (A-3) +... + 1 / (a-2004) (a-2005)

1.
a+1/a=2
a^2+1/a^2=(a+1/a)^2-2=4-2=2
a^3+1/a^3=(a+1/a)(a^2+1/a^2)-(a+1/a)=2*2-2=2
Conjecture a ^ n + 1 / A ^ n = 2
Proof: (mathematical induction)
Let n = k, that is, a ^ k + 1 / A ^ k = 2, a ^ (k-1) + 1 / A ^ (k-1)] = 2
a^(k+1)+1/a^(k+1)=[a^k+1/a^k](a+1/a)-[a^(k-1)+1/a^(k-1)]
=2*2-2=2
That is, n = K + 1 is also true
It is proved that a ^ n + 1 / A ^ n = 2 holds
two
1/a-1+1/(a-1)(a-2)+1/(a-2)(a-3)+...+1/(a-2004)(a-2005)
=1/a-1+[1/(a-2)-1/(a-1)]+[1/(a-3)-1/(a-2)]+...+[1/(a-2005)-1/(a-2004)]
=1/(a-2005)

8. There are two bulbs in series in the circuit, the power supply voltage is 12V, the voltage at both ends of bulb L1 is 9V, and the total resistance in the circuit is 30 ohm
8. There are two bulbs in series in the circuit, the power supply voltage is 12V, the voltage at both ends of bulb L1 is 9V, the total power of the circuit is 12V
The resistance is 30 ohm
(1) Voltage at both ends of bulb L2 (2) current in circuit (3) resistance of two bulbs

(1) U = U1 + U2, so U2 = u-u1 = 3V
（2）I=U/R=12V/30Ω=0.4A
(3) R1 / r2 = U1 / U2 = 3 / 1, R1 + R2 = 30 Ω, solving equations: R1 = 22.5 Ω, R2 = 7.5 Ω

How to solve an equation problem, 5x & # 178; + 5Y & # 178; = 65?
This problem is about the quadratic equation of one variable in volume one of grade nine.

5x²+5y²=65
x²+y²=13
The trajectory is a circle with a radius of √ 13

The factors that determine the resistance

The resistance is only related to these four factors
You can first find the same length and diameter of wire and copper wire, resistance measurement found that the resistance of copper wire is small
The resistance of two copper wires with the same length but different thickness was measured
The resistance of two copper wires with the same thickness but different length was measured
The lower the temperature, the smaller the resistance. Of course, if the temperature is low enough, its resistance is zero,
In addition, the resistance of glass increases with the decrease of temperature

1. To build a highway, it takes 80 days for engineering team a to contract alone, and 120 days for engineering team B to contract alone
① Now it is contracted by two engineering teams and can be completed in a few days
② If Party A and Party B have cooperated for 30 days, and Party A has another task, and Party B has to complete the rest of the work, how many days will it take to repair the road
2. For a project, it takes 9 days for Party A to complete it alone, 12 days for Party B to complete it alone, and 15 days for Party C to complete it alone. If Party A and Party C do it three days first, Party A leaves for some reason, and Party B takes over Party A's work. How many days will it take to complete five sixths of the project
3. For a project, it takes 20 days for Party A to complete it alone, and 25 days for Party B to complete it alone, so Party A will do it for 2 days first, and then how many days will it take for Party A and Party B to cooperate to complete the rest of the project
——————————Equation solution ~ thank you————————————————

1. To build a highway, it takes 80 days for Party A to contract alone, and 120 days for Party B to contract alone

The power of 800 w electrical appliances, how many times a point,
Heater,

1.25 hours

10000 / (square of 99 + 198 + 1)

10000 / (square of 99 + 198 + 1)
=10000 / (square of 99 + 99 * 2 * 1 + 1)
=Square of 10000 / (99 + 1) ^ 2
=Square of 10000 / 100 ^ 2
=1/10000
You see, do you understand? If not,
The most important thing is the method. If we master the method, we can solve similar problems!
Like this question oneself tries under, next time will be able to!
I wish you academic progress!