Eight matches can be used to make a rectangle or a square. Can eight matches be used to make a geometric figure with a larger area than the two figures?

regular octagon
Is larger than a rectangle or square

Use matches to put three triangles, five for two triangles and seven for three triangles. If you want to put 20 triangles, how many should you use

Three triangles need to be placed with matches, five triangles need to be placed in a row, seven triangles need to be placed in a row, and 41 triangles need to be placed if 20 triangles need to be placed
2×20+1
Good luck

How many ways can I swing a triangle with 12 matchsticks of the same length? Draw the triangle and mark the number of triangles on each side

2 5 5
3 4 5
3 5 4
4 3 5
4 5 3
4 4 4
5 2 5
5 3 4
5 4 3
5 5 2

Make four triangles of the same size with six identical matchsticks. How to make them? There should be figures

Make four triangles of the same size with six identical matchsticks,
It can only be found from space. It's a regular tetrahedron with six sides where the matchstick is

Let {an} satisfy A1 = 1. Sn = a1 + A2 + a3 + When n > = 2, find sn-sn-1
(2) Finding the general term formula an of sequence

1.
Sn=A1+A2+A3+…… +An=n^2
S(n-1)=A1+A2+A3+…… +A(n-1)=(n-1)^2
Sn-S(n-1)=n^2-(n-1)^2=2n-1
two
n> When = 2
An=Sn-S(n-1)=2n-1
The empirical calculation A1 = 1 also satisfies the above equation

It is known that the sum of the first n terms of the sequence an is Sn, and Sn + an = 1 / 2 (N2 + 5N + 2) (2 belongs to n *) to calculate A1 A2 A3 A4
It's just that I don't know how to work it out. I want to find a positive solution
What about bringing in SN,

When n = 1,
a1+a1=1/2(1*1+5*1+2)=4
a1=2
When n = 2
a1+a2+a2=1/2(2*2+5*2+2)
2+2*a2=8
a2=3
When n = 3,
a1+a2+a3+a3=1/2(3*3+5*3+2)
a3=4
When n = 4
a1+a2+a3+a4+a4=1/2(4*4+5*4+2)
a4=5

In the arithmetic sequence {an}, A3 = 8, S3 = 33, find the expression of the first n terms and TN of the sequence {an}
I already know the first n terms of the sequence {an} and Sn = - 3 / 2n ^ 2 + 31 / 2n
No nonsense!
Note that it's absolute

S3=A1+A2+A3
=A3+A3-D+A3-2D
33=3(8-D)
The tolerance is - 3, the first item is 8-2d = 14
So the general formula is an = 17-3n
The sum of the first n terms is, Sn = (31-3n) n / 2 (n = 1,2,3,4,5)
|A6+A7+...+AN|=-(A6+AN)(N-5)/2=(16-3n)(n-5)/2
SN=(3n^2-31n+160)/2 (n>5)

{an} is an arithmetic sequence, and A1 = 2, a1 + A2 + a3 = 12 (1) find the general term formula of the sequence {an} (2) let BN = an * 2 ^ an, find the sum TN of the first n terms of the sequence {BN}

(1)S3=3a2=12
a2=4
d=a2-a1=2
an=2n
(2)bn=2n*2^2n=2n*4^n
Tn=2*4^1+4*4^2+6*4^3+… +2(n-1)4^(n-1)+2n*4^n
4Tn=2*4^2+4*4^3+6*4^4+… +2(n-1)4^n+2n*4^(n+1)
3Tn=-2*4+-2(4^2+4^3+… +4^n)+2n*4^(n+1)
=-8-2(3\4^(n+1)-32)+8n*4^n
=-8-3\(4^n-64)+8n*4^n
It should be easier to simplify here
Tn=3\[-8-3\(4^n-64)+8n*4^n]

An is an arithmetic sequence. A3 = - 6, A6 = 0. Find the sum of the first n terms of the absolute values of an and an and the formula TN

a3=a6+3d
d=2
a1=a3-2d=-10
an=2n-12
Tn=n(11-n) (n6)

Given that the sequence an is an equal ratio sequence, and A1 = 1, A4 = - 27, find the general term formula of the sequence an

The basic formula of equal ratio sequence: an = A1 * q ^ (n-1), q is the common ratio, n is the nth term
a4=a1*q^(4-1) → 27=1*q^3 → q^3=27 → q=27^1/3=3,
So an = 3 ^ (n-1) is the general formula of an

Eight matches can be used to make a rectangle or a square. Can eight matches be used to make a geometric figure with a larger area than the two figures?