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