Proof: n ^ 3 + 3 / 2n ^ 2 + 1 / 2n-3 is an integer divisible by 3 for any natural number n

Proof: n ^ 3 + 3 / 2n ^ 2 + 1 / 2n-3 is an integer divisible by 3 for any natural number n

Mathematical induction first proves that n = 0 can be divisible, and then assumes that n = k is feasible, proving that n = K + 1 is still feasible: K ^ 3 + 3 / 2K ^ 2 + 1 / 2k-3 is known to be feasible, K ^ 3 + 3K ^ 2 + 3K + 1 + 3 / 2K ^ 2 + 3K + 3 / 2 + 1 / 2K + 1 / 2-3 is known to be feasible, K ^ 3 + 3K ^ 2 + 3K + 3 / 2K ^ 2 + 3 / 2 + 1 / 2K + 1 / 2-3 is simplified, which is "the above hypothesis plus three" plus a bunch of terms with coefficient three (constant term is eliminated)
When calculating the electric energy consumed by electrical appliances, what is the time unit
In seconds
Formula w = uit
(U is voltage (V / V), I is current (A / a), t is time (s / s))
Where is the unit of time
W = Pt, the unit of P is kW, the unit of time is h, the unit of W is kwh; the unit of P is w, the unit of time is s, and the unit of W is J
How much is 2-root plus 3 root plus 2 root plus 3 root
The original formula ^ 2 = 2 + √ 3 + 2 - √ 3 + 2 × (√ (2 + √ 3) ×√ (2 - √ 3))
=4+2×1
=6
So the original formula = √ 6
Double root 2
Let √ (2 - √ 3) + √ (2 + √ 3) = a
Both sides square at the same time
a²＝6
∴a＝√6
√(2-√3)=√[(4-2√3)/2]=√[(√3-1)^2/2]=(√3-1)/√2
Similarly, √ (2 + √ 3) = (√ 3 + 1) / √ 2
Therefore, the original formula = (√ 3-1 + √ 3 + 1) / √ 2 = (2 √ 3) / √ 2 = √ 2 * √ 3 = √ 6
There are four different natural numbers, in which the sum of any two numbers can be divided by 2, and the sum of any three numbers is a multiple of 3
From the sum of any two numbers can be divided by 2, we can know that they are either all odd numbers or all even numbers. From the sum of any three numbers are all multiples of 3, if they are all even numbers, all four numbers are multiples of 6. If they are all odd numbers, the difference between any two numbers must be multiples of 6
From the fact that the sum of any two numbers can be divided by 2, we can see that all of them are congruent to 2; from the fact that the sum of any three numbers is a multiple of 3, we can see that all of them are congruent to 3; that is, the difference between any two numbers must be a multiple of 6. Generally speaking, as long as the difference between any two numbers is a multiple of 6, it can meet the requirements of the topic, such as: 0, 6, 12, 18; 1, 7, 13, 19; 2, 8, 14, 20, 3, 9, 15, 21, etc
From the fact that the sum of any two numbers can be divided by 2, we can see that all of them are congruent to 2; from the fact that the sum of any three numbers is a multiple of 3, we can see that all of them are congruent to 3; that is, the difference between any two numbers must be a multiple of 6. Generally speaking, as long as the difference between any two numbers is a multiple of 6, it can meet the requirements of the topic, such as: 0, 6, 12, 18; 1, 7, 13, 19; 2, 8, 14, 20, 3, 9, 15, 21, etc
3 9 15 21 and so on
2 4 6 8
The common unit of electric energy in life is () (kW &; H), which is the electric energy consumed by the electric appliance with electric power of ()
Kilowatt hour one kilowatt hour
What's the area of an equilateral triangle? Its side length is 6. I don't want the formula, just the answer
9 times root 3
Why can any natural number be divided by 3 if the sum of each digit is a multiple of 3
For example, 67 is (6 + 7) / 3 = 4, so 67 can be divided by 3
Why can any natural number be divided by 3 if the sum of each digit is a multiple of 3
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Let's first look at two digit numbers, such as the combination of numbers
A + B is a multiple of 3
Then 10 * a + B = 9A + (a + b)
9A is divisible by 3, a + B is divisible by 3, so 10 + B is divisible by 3
Look at the three digit, such as digital ABC combination
A + B + C is a multiple of 3
Then 100 * a + 10 * B + C = 99A + 9b + (a + B + C)
99A, 9B, (a + B + C) can be divisible by 3, so 100 * a + 10 * B + C can be divisible by 3
In fact, for any natural number a (1) a (2) a (3) a (4). A (n)
If a (1) + a (2) + a (3) +... + a (n) is a multiple of 3
that
a(1)*10^(n-1)+a(2)*10^(n-2)+.+a(n-1)*10+a(n)
=a(1)*[10^(n-1)-1]+a(2)*[10^(n-2)-1]+...+a(n-1)*9+[a(1)+a(2)+...+a(n)]
Every item in the middle can be divided by 3
“a...bcde”
= e + 10d + 100c + 1000b + ... + 10...00a
= ( e + d + c + b + ... + a ) + [ 9d + 99c + 999b + 9...99a ]
Because the sum in is divisible by 3,
So if we divide the sum of () and () by 3,
Then "a... BCDE" must be divisible by 3 (and vice versa).
(6 + 7) / 3 = 4? Wrong calculation
What is the difference between electric power and electric work
Electric work refers to the amount of electric energy consumed, w = uit, in joules and kilowatt hours
Electric power refers to the speed of power consumption, P = UI. The units are w and kW
The relation between them is p = w / T
Area formula of equilateral triangle
We know that the side length of an equilateral triangle is 3cm
Q: what is the area of this triangle in square centimeters
I don't know if you have studied trigonometric function!
S=1/2*L*L*sin60
=9√3/4
If you don't learn,
S=1/2*L*H
=9√3/4
three point eight nine
Three point nine
s=(√3/4)a^2
=9/4*√3=3.9
The title is wrong!
apply a formula
It's 4.5 square centimeters
Proving 1 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + by mathematical induction +The remainder of ^ 2 is divisible by 1-5n
The remainder is 0 divided by 31
When n = 1
1+2^1+2^2+2^3+4^2=1+2+4+8+16=31
establish
Let K be established
1+...+2^(5k-1)=31p
When K + 1
1+..+2^(5k-1)+2^5k+2^(5k+1)+2^(5k+2)+2^(5k+3)+2^(5(k+1)-1)
=31p+2^5k(1+2^1+2^2+2^3+2^4)
=31(p+2^5k)
K + 1
The original formula + 1 = 2 ^ 5N
That is, the original formula = 2 ^ 5n-1
It's easy to prove it again
1. When n = 1, 2 ^ 0 2 ^ 1 + 2 ^ 4 = 31. So the remainder is 0. 2. Suppose that the conjecture holds when n = K (K ＞ = 1), then there are 2 ^ 0 2 ^ 1 2 ^ 2 Then when n = K + 1, there are 2 ^ 0 2 ^ 1 2 ^ 2 + 2 ^ (5 (K 1) - 1) = 2 ^ (5... Expansion
1. When n = 1, 2 ^ 0 2 ^ 1 + 2 ^ 4 = 31. So the remainder is 0. 2. Suppose that the conjecture holds when n = K (K ＞ = 1), then there are 2 ^ 0 2 ^ 1 2 ^ 2 Then when n = K + 1, there are 2 ^ 0 2 ^ 1 2 ^ 2 + 2 ^ (5 (K 1) - 1) = 2 ^ (5 K 5) - 1 = 2 ^ (5 K) * 32 - 1 = (31 M 1) * 32 - 1 = 31 * (32 m 1) Stow