High school math problems, masters, please come in, thank you

Example 1: Sum(k=1 n) k

k+1)³-k³=3k²+3k +1

Therefore k = (1 3)[(k+1) -k -3k-1].

Use k = 1, 2, 3 ,..n in turn to get the following:

n²=(1/3)[(n+1)³-n³-3×n-1]

Add the above n equations to get :

k=1→n)∑k²=(1/3)[(n+1)³-1-3(1+2+3+..n)-n}=(1/3)[(n+1)³-1-3(1+n)n/2-n}

1/3)[(n+1)³-3n(n+1)/2-(n+1)]=(1/3)(n+1)[(n+1)²-3n/2-1]

1/3)[2(n+1)²-3n-2]/3=(1/6)(n+1)(2n²+n)=n(n+1)(2n+1)/6

Example 2Sum (k=1 n) 1 (k +3k+2).

1/(k²+3k+1)=1/(k+1)(k+2)=1/(k+1)-1/(k+2)

k=1→n)∑1/(k²+3k+2)=[1/2-1/3]+[1/3-1/4]+[1/4-1/5]+.1/(n+1)-1/(n+2)]

1/2-1/(n+2)=n/2(n+2)

I can't see clearly with my phone, I'm depressed! If you're going to break it down, multiply it by a half of an a.

You add mine, and I'll tell you.

Vector om and vector on = 2x+y

The graph enclosed by x 3, x-y+6 0, x+y 0 has 2x+y, and the minimum value of 2*(-3)+3=-3; is obtained at the intersection of x-y+6=0, x+y=0 (-3,3);

2x+y obtains the maximum value of 2*3+9=15 at the intersection of x=3,x-y+6=0 (3,9).

The value range of the vector om and vector on is [-3,15].

Answer: greater than or equal to 0 less than or equal to 15 root number 2 (correct question vector om vector on). Idea: Draw a straight line of three equations about the point n on the plane coordinate axis, i.e., x-y+6=0, x+y=0, x=3. It was judged that the n area was determined....

I guess it's om*on.

After the vector is represented by coordinates, its inner product is directly the sum of the product of the corresponding coordinates, that is: om*on=2x+y

After drawing the three straight lines x=3 x-y+6=0 x+y=0 in the Cartesian coordinate system, the area that meets the requirements is the area enclosed by these three straight lines. It is not difficult to see that this is essentially a linear programming problem. Within the specified range, the line 2x+y=z can be moved to the left to the intersection of the two straight lines x-y+6=0 and x+y=0, that is:

3,3) Z=-3 at this point

In the process of moving the line 2x+y=z to the right, because the absolute value of its slope is greater than the absolute value of the slope of the line x-y=0, 2x+y=z will pass through the intersection point [3,9] of the two lines x-y+6=0 x=3 when it is on the far right, and z=15 at this time

Therefore, when 2x+y=z moves from left to right, the value from 3 to 15 will be taken to the range: [3,15].

72+18 times root number 3

1) Side. When the ball moves to the point where it touches the inner walls of both sides of the triangular prism container at the same time, the point of contact between the ball and the inner wall of the two sides must be some distance from the edges of the inner walls of both sides, and the distance between the two contact points and the edges is equal. From the contact point to the prism as a perpendicular line, and then connect the center of the ball and the perpendicular point, the center of the ball and the contact point, so as to form a right triangle.

The distance from the contact point to the edge: the radius of the ball tan60° = 1 3 = 3 The area of the inner wall at this distance is: distance Ridge length = 3 4 3 = 12The area of the inner wall on both sides is: 12 2 = 24

A straight triangular prism has three edges, and the total inaccessible area is: 24 3 = 72 (cm).

2) Underside. The tangent area of the sphere to the base is only a regular triangle with a side length of 2 times the root number 3, and the remaining 9 times the root number 3 times times 2 faces.

It is 72+18 times the root number 3

Let's assume that the container is as small as possible until the ball is just right to get stuck inside, i.e., it is tangent to each side.

This should be a triangular prism with a base side length of 2 cm and a height of 2 cm, calculate its surface area.

Then calculate the surface area of the original triangular prism, and find the difference, which is the area that cannot be touched.

That miniature version is actually the 6 corners of the space that can't be touched.,The landlord can draw a picture to ponder.,I think this is what I drew and can't tell.,To space imagination.。

pa face abc, so pa bc, and bca=90°, so bc acpa ac=a, so bc face pac

Take the midpoint F of the PC and connect DF and AF, DF is the median line, so DF BCBC plane PAC, so DF plane PAC

So daf is the angle between ad and the face pac.

Let Pa=ab=2, and in the right-angle abc BC=1, AC= 3DF=1, 2BC=1, 2, AD= 2

So sin daf=df ad= 2 4

3) The presence of e is such that the dihedral angle a-de-p is a straight dihedral angle.

Do de pb over to e and connect ae through d. This point is that the condition is satisfied, because Pab is an isosceles right angle, so ad pd, and pd de, ad de=d

So pd face ade, pd face pde, so face pde face ade so dihedral angle a-de-p is a straight dihedral angle.

Require the position of point E, bc=1 pc= 7 pb=2 2 pd= 2cos dpe=pd pe=pc pb

So pe=4 7 7

pe/pc=4/7

That is, e is at a distance of 4 7 from point p.

1.PA Bottom ABC

PA BC and BC AC

BC planar pac

2.de bc, and d is the midpoint of pb.

de face pac

sin angle dae=de ad=1 2bc ad, ad=root2 2 ab=root2 bcsin, angled dae=1 2bc root2 bc=root2 divide by 43does not exist, because to meet this must ad,ae mean pde, there is only one perpendicular line of the surface that passes the same point, so it is impossible to exist at the same time ad,ae perpendicular, in other words, there can be no point e so that the dihedral angle a-de-p is a straight dihedral angle.

Do poor, seek derivation, and then use monotonicity. That's the idea.

Because o is the outer center, the projection of o on each side is precisely the midpoint of each side, and all ao*ab=|ao|*|ab|*cos∠oab=(|ao|*cos∠oab)*|ab|=1/2*|ab|=9 2, the same gives ao*ac=1 2*|ac|=8, multiply Ab and AC respectively on both sides of ao=xab+yac, and let ab*ac=k, then.

9/2=9x+ky；

8=kx+16y；

x+2y=1, the solution is x=1 10, y=9 20, k=8, and is given by k=ab*ac=|ab|*|ac|*cos∠bac=12cos∠bac=8

Get cos bac=2 3.

Hello classmates