How many figures can four matches make What happened? say，say。

24, really

It takes 5 matches to make a figure, 8 matches to make two figures together, and 12 matches to make three figures together,
It takes 15 matches to put four figures together, so how many matches does the nth figure need?
It takes 19 matches to put five figures together

When n is odd, the nth graph needs [5 + 7 (n-1) / 2] matches
When n is even, the nth graph needs [8 + 7 (n-2) / 2] matches

How many different patterns with triangles can six matchsticks make?

Different angles are different figures. For example, if two vertices coincide, one triangle is fixed and the other rotates around the vertex, several patterns will be obtained

How many matches do n figures need
There are many different ones on the Internet. It seems that they are all wrong. It's better to list only one formula instead of odd and even numbers. By the way, can matches be 63 and 70? Why

It's better to divide the parity

Given the infinite arithmetic sequence {A & nbsp; n}, in the first n terms and Sn, S6 ＜ S7, and S7 ＞ S8, then ()
A. In the sequence {an & nbsp;}, a7 & nbsp; is the largest B. in the sequence {an}, A3 or A4 is the largest C. The sum of the first three terms S3 must be equal to the sum of the first 11 terms S11. D. when n ≥ 8, an < 0

From S6 ＜ S7, and S7 ＞ S8, a1 + 6D ＞ 0, a1 + 7d ＜ 0, a7 ＞ 0, a8 ＜ 0, D ＜ 0. So when n ≥ 8, a8 ＜ 0

SN is the first n term of the arithmetic sequence {an} and S5 is less than S6, S6 = S7 > S8
1) The tolerance DS5 (3) A7 = 0 (4) S6 and S7 are the maximum values of Sn
What's wrong with the above?

1.
A6=S6-S5>0
A7=S7-S6=0
A8=S8-S7

Let an be an arithmetic sequence, Sn be the sum of its first n terms, and s5s8, then the following conclusion is wrong
A. Ds8, d.s6 and S7 are the maximum values of Sn
Please state the reason or the solution?

C from the meaning of the title, we can see that DA2 > A3 > A4 > A5 > A6 > A7 = 0 > A8 > A9 > of the arithmetic sequence an
>an
So S6 = S7 is the maximum of Sn
s9

In the arithmetic sequence {an}, Sn is the sum of its first n terms, and s6s8
A1 is the biggest of all, right? Why

Yes
Let Sn = an ^ 2 + BN, then a = D / 2
∵ Sn increases on [1,7] and decreases on [7, + ∞]
∴A=d/2

It is known that SN is the sum of the first n terms of the arithmetic sequence an, and satisfies S6 > S7 > S5,
Please judge whether it is correct and give reasons
(1) Tolerance DS3; (3) S11 > 0; (4) S8 > S6; (5) S13

Sn=n[2A1+(n-1)d]/2
S5=5(2A1+4d)/2=5A1+10d
S6=6(2A1+5d)/2=6A1+15d
S7=7(2A1+6D)/2=7A1+21d
S6>S7,6A1+15d>7A1+21d,A1S5,7A1+21d>5A1+10d,A1>-5.5d
-5 d 0 is right
S11=11(2A1+10d)/2=11A1+55d>-11*5.5d+55d>0
(4) S8 > S6 is wrong
S8=8(2A1+7d)/2=8A1+28d
S8-S6=8A1+28d-(6A1+15d)=2A1+13d

The sum of the first n terms of the arithmetic sequence {an} is SN. If a1 + A9 + a11 = 30, then the value of S13 is ()
A. 65B. 70C. 130D. 260

If the tolerance is D, because a1 + A9 + a11 = 30, ｛ 3A1 + 18D = 30, ｛ A7 = 10, then S13 = 13 (a1 + A13) 2 = 13a7 = 130, select C

How many figures can four matches make What happened? say，say。