Let P1, P2, P3 ~ PN The reverse order number of is k, then What is the reverse order number of PN ~ P3, P2, P1?

Let P1, P2, P3 ~ PN The reverse order number of is k, then What is the reverse order number of PN ~ P3, P2, P1?

t=(n-1)*n/2+k
Because the reverse order number of the following column is (n-1) * n / 2
The reverse order number of the following line is the same as that of the above one, and the constant is K
First of all, let's assume that the reverse order numbers of the elements in P1, P2,... PN are T1, T2 TN
That is to say, the reverse number of P1 is T1 (in fact, T1 = 0, which is written as T1 for the convenience of illustration), and the reverse number of P2 is T2 The reverse number of PN is TN
In addition, from the definition of inverse ordinal number, we can know that P1, P2,... PN are different numbers
For the next step of reasoning, we first explain a conclusion, that is, two adjacent numbers are exchanged once. If the former number is larger than the latter number, then the reverse order number of the whole sequence is reduced by 1, otherwise it is expanded by 1
First of all, let's assume that the reverse order numbers of the elements in P1, P2,... PN are T1, T2 TN
That is to say, the reverse number of P1 is T1 (in fact, T1 = 0, which is written as T1 for the convenience of illustration), and the reverse number of P2 is T2 The reverse number of PN is TN
In addition, from the definition of inverse ordinal number, we can know that P1, P2,... PN are different numbers
For the next step of reasoning, we first explain a conclusion, that is, two adjacent numbers are exchanged once. If the former number is larger than the latter number, then the reverse order number of the whole sequence is reduced by 1, otherwise it is increased by 1
That is to say, for the sequence a, B → B, a
If a > b, then the inverse ordinal number will change from 1 to 0, and vice versa
In this way, it is easy to know that for an element PN in the sequence P1, P2,... PN, its reverse order number is TN, which means that in P1, P2,... PN-1, the number of numbers larger than PN is TN, and the number of numbers smaller than PN is n-1-tn
Then the transformation of P1, P2,... PN to PN, P1, P2,... PN-1 needs to go through n-1 times of adjacent transformation, and if considered as a whole, there are TN times of inverse number minus 1 transformation and n-1-tn times of inverse number plus 1 transformation, so the inverse number of PN, P1, P2,... PN-1 is k-tn + n-1-tn
In the same way, for PN-1, for the sequence PN, P1, P2,..., PN-1, in PN, P1, P2,..., pn-2, there are TN-1 numbers larger than that and n-2 numbers smaller than that, so PN, P1, P2,..., PN-1 is transformed into PN, PN-1, P1, P2,..., PN-1, ... pn-2, need to go through n-2 times of transformation, and overall consideration, to go through TN-1 times of inverse number minus 1 transformation, there is n-1-tn-1 times of inverse number plus 1 transformation, so PN, PN-1, P1, P2,... Pn-2's inverse number becomes k-tn + n-1-tn-tn-1 + n-2-tn-1
……
And so on, until PN,... P2, P1, we can prove that its inverse ordinal number is k-tn + n-1-tn-tn-1 + n-2-tn-1-tn-2 + n-3-tn-2 - -t1+0-t1=k-2(tn+tn-1+…… +t1)+n-1+n-2+…… +1
We know TN + TN-1 + from the known conditions +t1=k
So the reverse order number of the transformed PN,... P2, P1 is k-2k + (n-1) n / 2 = (n-1) n / 2-k
As shown in the figure, in the triangular pyramid p-abc, PA = Pb = PC = BC, and the angle BAC = π / 2, then the projection of point P on the ground ABC is triangular__ heart
(that's the dichotomy)
External
Is the intersection of the vertical bisectors of each side
Verification: P1 ^ 1 + 2 * P2 ^ 2 + 3 * P3 ^ 3 +... N * PN ^ n = P (n + 1) ^ (n + 1) - 1. (n ∈ n *)
Prove by mathematical induction
1, when n = 1
P(1,1)=1
P(2,2)-1=2*1-1=1
P (1,1) = P (2,2) - 1 holds
2. Suppose n = K and K belongs to n,
That is, P1 ^ 1 + 2 * P2 ^ 2 + 3 * P3 ^ 3 +... K * PK ^ k = P (K + 1) ^ (K + 1) - 1 holds
Then when n = K + 1
Left = P1 ^ 1 + 2 * P2 ^ 2 + 3 * P3 ^ 3 +... + k * PK ^ k + (K + 1) P (K + 1) ^ (K + 1)
=P(k+1)^(k+1)-1+(k+1)P(k+1)^(k+1)
=(k+1)!-1+(k+1)*(k+1)!
=(k+1)!(k+2)-1
=(k+2)!-1
=P(k+2)^(k+2)-1
=Right
Don't understand, please hi me
Proof: when n = 1, that is, left = P1, right P2 ^ 2-1
What's missing from the title? Follow up: no leakage
As shown in the figure, in the triangular pyramid p-abc, ab ⊥ BC, ∠ BAC = 30 °, BC = 5, and PA = Pb = PC = AC. then the distance from point P to plane ABC is______ .
Because PA = Pb = PC, then their projections in plane ABC are equal, P in plane ABC should be at the outer center of triangle ABC, and triangle ABC is a right triangle, so the outer center should be at the midpoint D of the hypotenuse, PD ⊥ bottom ABC, ∠ BAC = 30 °, AC = 2BC = 10, BD = 102 = 5, Pb = AC = 10, triangle PBD is a right triangle
It is known that the images of the first-order function y = x + m and the inverse scale function y = m + 1 / X (M is not equal to - 1) pass through the point P (x0,3) of the first quadrant. The analytic expressions of the two functions are obtained
3=x0+m
x0=3-m
Substituting 3 = (M + 1) / x0, we get the following result:
3(3-m)=m+1
9-3m=m+1
M=2
The analytic expressions of two functions are as follows
y=x+2
y=3/x
Let 3 = x + m, and 3 = m + 1 / X solve the system of equations to get x = 1, M = 2; or x = - 1
Take: x = 1, M = 2. Substituting M = 2 into y = x + m, y = x + 2, substituting M = 2 into y = m + 1 / x, y = 2 + 1 / X
In the sharp triangular pyramid p-abc, the angle BAC = 90 degrees, PA = Pb = PC = BC = 2Ab = 2, the verification plane PBC, the vertical plane ABC 2, the two faces angle
In the sharp triangular pyramid p-abc, the angle BAC = 90 degrees, PA = Pb = PC = BC = 2Ab = 2, the verification plane PBC, the vertical plane ABC 2, and the cosine value of the two faces b-ap-c
Make PD ⊥ BC and hand BC to d
Pb = PC = BC, △ PBC is positive
D is the midpoint of BC, ad = √ 3
Ad is the middle line of RT △ ABC hypotenuse, ad = BC / 2 = 1
AD^2+PD^2=PA^2
Δ pad is RT △ PD ⊥ ad
PD ⊥ face ABC, face PBC ⊥ face ABC
Make be ⊥ PA hand over PA to e
Sin θ = H / be = (3V / s) / be
Given that the images of the first-order function y = x + m and the inverse scale function y = m + 1 / X (M is not equal to - 1) pass through the point P (x0,3) of the first quadrant, the value of x0 is calculated
Substituting (x0,3) into: 3 = x0 + m, 3 = m + 1 / X0
The solution is x0 = 1
In the triangle ABC, ab = AC, P is a point on BC. It is proved that the square of AB = the square of PA + Pb * PC
It can be proved when p is at the midpoint of BC
Because P is the midpoint of BC, ab = AC, BP = PC AP ⊥ BC
So the angle APC is equal to 90 ° BP times PC = BP & sup2; using Pythagorean theorem AB & sup2; = BP & sup2; + AP & sup2; + Pb & sup2; + BP times PC
I don't know if it's the first day
It is known that the intersection point of the first-order function y = x + m and the inverse scale function y = m + 1 / X (M is not equal to - 1) in the first quadrant is p (x1,3)
1. Find point P
2. Finding the expression of first order function and inverse proportion function
A:
（1）
According to the meaning of the title:
3=x1+m
3=(m+1)/x1
The solution is X1 = 1, M = 2
So point P is (1,3)
(2) The linear function is y = x + 2, and the inverse proportion function is y = 3 / X
In the triangle ABC, ab = AC = 6, P is any point on BC, please use the learned knowledge to try to find the value of PC times Pb + PA quadratic
Drawing is available
∵AB=AC
∴∠B=∠C
∴cosB=cosC
Available cosine theorem
PA²=36+PB²-12PBcosC
-12cosC=(PA²-36-PB²)/PB
PA²=36+PC²-12PCcosB
-12cosB=(PA²-36-PC²)/PC
(PA²-36-PB²)/PB=(PA²-36-PC²)/PC
PC×PA²-36PC-PC×PB²=PB×PA²-36PB-PB×PC²
(PC-PB)PA²-(PC-PB)36-(PB-PC)PCPB=0
PC×PB+PA²=36
If we regard this triangle as an equilateral triangle and P as the midpoint of BC, we can get the value 108 + 54 * (3) ^ (1 / 2)
PA²=36+PB²-12PBcosC
-12cosC=(PA²-36-PB²)/PB
PA²=36+PC²-12PCcosB
-12cosB=(PA²-36-PC²)/PC
(PA²-36-PB²)/PB=(PA²-36-PC²)/PC
PC×PA²-36PC-PC×PB²=PB×PA²-36PB-PB×PC²
(PC-PB)PA²-(PC-PB)36-(PB-PC)PCPB=0
PC×PB+PA²=36
Over the point Ao ⊥ BC. ∵AC=AC=6 ∴BO=CO
The triangle apo is a right triangle and the triangle AOC is a right triangle
PC×PB+PA²
=（CO+OP)(CO-OP)+PA²
=CO²-OP²+PA²
=PA²-OP²+CO²
=AO²+CO²
=AC²
=36