Can you make four triangles of the same size with six matches? If so, please name your figure emergency

Can you make four triangles of the same size with six matches? If so, please name your figure emergency

Three dimensional consideration can be placed into a triangular pyramid with only six edges

Can you use six matches of the same length to form four triangles of the same size? What shapes can they form?

A tetrahedron whose four sides are equilateral triangles

Use eight matchsticks to make two squares and eight triangles

What do you want upstairs
answer:
Square: one for every four
Triangle: a figure formed by folding two squares and rotating one of them 45 degrees

It takes 7 matches to make two squares, 12 matches to make four squares, 17 matches to make six squares, 52 matches to make ten squares, and how many matches to make n squares?

Let n (n is even) squares have s matches in total
When n = 2, s = 7 = 5 * 2 / 2 + 2
When n = 4, s = 12 = 5 * 4 / 2 + 2
When n = 6, s = 17 = 5 * 6 / 2 + 2
When n = 8, s = 22 = 5 * 8 / 2 + 2
When n = 10, s = 27 = 5 * 10 / 2 + 2
.
So when you put n squares together, you need matches: 5N / 2 + 2 (pieces)
(the original title "spelling 10 to 52" should be a transcription error)

Given that the tolerance is greater than 0 of the arithmetic sequence {1 / an}, satisfy a2a4 + a4a6 + a6a2 = 1, A2, A4, a8 in turn into a proportional sequence, find the general formula of an

Let the tolerance be K
Then an = A0 + K (n-1)
So a2a4 + a4a6 + a6a2 = (a1 + k) (a1 + 3K) + (a1 + 3K) (a1 + 5K) + (a1 + 5K) (a1 + k)
The equation: (a1 + k) (a1 + 3K) + (a1 + 3K) (a1 + 5K) + (a1 + 5K) (a1 + k) = 1
A2, A4 and A8 are in equal proportion sequence
So A2 / A4 = A4 / A8
So A4 ^ 2 = a2a8
So (a1 + 3K) ^ 2 = (a1 + k) (a1 + 7K)
The two equations contain two unknowns, so they can be solved simultaneously
The final solution is A1 = k = 11 / 22 under the root
So the general formula is: an = 11 / 22 under N root

The tolerance of the arithmetic sequence {an} is 2. If A2, A4 and A8 are equal ratio sequence, then the first n terms of {an} and Sn = ()
A. n（n+1）B. n（n-1）C. n(n+1)2D. n(n−1)2

From the meaning of the title, we can get A42 = A2 · A8, that is, A42 = (a4-4) (A4 + 8), the solution is A4 = 8, ∧ A1 = a4-3 × 2 = 2, ∧ Sn = Na1 + n (n − 1) 2D, = 2n + n (n − 1) 2 × 2 = n (n + 1), so we choose a

What is the tolerance of the arithmetic sequence whose general formula is an = 3n-2?

3
What's the coefficient in front of N? That's the tolerance

It is known that the general term formula of arithmetic sequence {an} is an = 2N-1, so Sn can be obtained
It's an = 2n + 1

a1=2-1=1
So Sn = (1 + 2n-1) n / 2 = n & # 178;

Given the arithmetic sequence [an], Sn = [(an + 1) / 2] ^ 2, find the general term formula of an
N is subscript, please help!

Because Sn = [(an + 1) / 2] ^ 2
So s (n-1) = [(a (n-1) + 1) / 2] ^ 2
By subtracting the two formulas, 4An = (an + 1) ^ 2 - (a (n-1) + 1) ^ 2 is obtained
It is reduced to (a (n-1) + 1) ^ 2 = (an-1) ^ 2
So a (n-1) + 1 = an-1
So an = a (n-1) + 2
Because S1 = [(a1 + 1) / 2] ^ 2
So A1 = 1
So an = a1 + 2 (n-1) = 2N-1

Given the general formula of arithmetic sequence {an} an = 2n + 1, find SN

Sn=(a1+an)*n/2
=(3+2n+1)*n/2
=(n+2)n