The inequality ln (1 / N + 1) > 1 / N ^ 2-1 / N ^ 3 is proved for any positive integer n

The inequality ln (1 / N + 1) > 1 / N ^ 2-1 / N ^ 3 is proved for any positive integer n

F (x) = x ^ 2 + AlN (1 + x), let a = - 1, G '(x) = x ^ 3-x ^ 2 + ln (1 + x), then G' (x) = [x ^ 3 + (x-1) ^ 2] / (1 + X), when x > 0, G '(x) > 0 holds, then G (x) increases monotonically on (0, + ∞), so there must be g (x) > G (0) = 0 and 1 / N ∈ (0,1), so let x = 1 / N also holds, so
Physical formula, electric heating, electric power
1. Electric power:
(1) , w = uit = Pt = UQ (universal formula)
(2) W = i2rt = u2t / R (pure resistance formula)
2. Electric power:
(1) , P = w / T = UI (universal formula)
(2) , P = I2R = U2 / R (pure resistance formula)
3. Joule's Law:
(1) , q = i2rt (universal formula)
(2) , q = uit = Pt = UQ, electric quantity = u2t / R (pure resistance formula)
Power P = UI
U voltage
I current
Electric heating rate: W1 = I ^ 2 * r
Electric power: W2 = UI
When the load is pure resistance, the heating rate and electric power are equal
When the load is not pure resistance (for example, the electric fan converts electric energy into wind energy as well as heat energy), the electric heating rate is not equal to the electric power
If the side length of an equilateral triangle is x, what is the functional relationship between its area y and X?
Y = x * xsin60 ° / 2 = root 3x & sup2 / / 4
Y = (x root 3) / 4
y=sin60°* x * x ÷2
Prove by mathematical induction: ln (1 + 1 * 2) + ln (1 + 2 * 3) + +Ln [1 + n (n + 1)] > 2n-3 (n belongs to n *)
RT
NLN [n ^ 2] = 2lnn > 2, when n > 2
Therefore, when n + 1, the proposition still holds
By induction, the original proposition always holds
Deformation formula of physical electric power formula under different conditions
Please write down the situation
(1) series circuit P (electric power) U (voltage) I (current) w (electric work) r (resistance) t (time) current is equal everywhere I1 = I2 = I total voltage equals to the sum of voltage at both ends of each electrical appliance u = U1 + U2 total resistance equals to the sum of resistance R = R1 + R2 U1: U2 = R1: R2 total electric work equals to the sum of electric work w = W1 + W2
p=UI=I^2R=Q/t=u^2/R
If the hypotenuse of an isosceles right triangle is 2xcm long, its area is y square centimeter
1> Find the function expression of Y with respect to x, and find out the value range of X
If the length of the hypotenuse of an isosceles right triangle is 2xcm, then the height of the hypotenuse is also the center line of the hypotenuse. If the length is x, then
The value range of area y = 1 / 2 * 2x * x = x ^ 2 x: x > 0
Prove 1 + 2 + 3 + by mathematical induction +N 2 = n 4 + n 22, then when n = K + 1, the left end should be added on the basis of n = K______ .
When n = k, the left end of the equation is 1 + 2 + +K2, when n = K + 1, the left end of the equation = 1 + 2 + +k2+（k2+1）+（k2+2）+（k2+3）+… +(K + 1) 2, added 2K + 1 term, namely (K2 + 1) + (K2 + 2) + (K2 + 3) + +So the answer is: (K2 + 1) + (K2 + 2) + (K2 + 3) + +（k+1）2
Joule's law is the same as the formula for calculating electric power. What's the difference?
It's different!
Joule's Law: q = I & # 178; RT
Electric power: P = I & # 178; R
The difference between the two is: is it in pure resistance circuit
Joule's law is a law that quantitatively explains that conduction current converts electric energy into internal energy.
Non pure resistance circuit: q = I ^ 2rt
If the length of hypotenuse of isosceles right triangle is 2xcm, its area is YCM & sup2;
Find the function relation of Y with respect to x, and find the value range of X
Quadratic function
If the length of the hypotenuse of an isosceles right triangle is 2xcm, the right side = the length of the hypotenuse / (√ 2) = 2x / (√ 2) = x √ 2 (CM),
Y = right angle side & sup2 / 2 = (x √ 2) & sup2 / 2 = x & sup2;,
That is y = x & sup2;, x > 0
Prove 1 + 2 + 3 + by mathematical induction +N 2 = n 4 + n 22, then when n = K + 1, the left end should be added on the basis of n = K______ .
When n = k, the left end of the equation is 1 + 2 + +K2, when n = K + 1, the left end of the equation = 1 + 2 + +k2+（k2+1）+（k2+2）+（k2+3）+… +(K + 1) 2, added 2K + 1 term, namely (K2 + 1) + (K2 + 2) + (K2 + 3) + +So the answer is: (K2 + 1) + (K2 + 2) + (K2 + 3) + +（k+1）2