5. Find the rules and continue to draw ○△○△○□（ ,）（ ,）（ ,）（ .） □○○□○○（ ,）（ ,）（ ,）（ .） ○△○○△○○（ ,）（ ,）（ ,）（ .）

5. Find the rules and continue to draw ○△○△○□（ ,）（ ,）（ ,）（ .） □○○□○○（ ,）（ ,）（ ,）（ .） ○△○○△○○（ ,）（ ,）（ ,）（ .）

○△○△○□（○）（△）（○）（△）（○） （□）
□○○□○○（□）（○）（○）（□） （○） （○）
○△○○△○○（○）（△）（○）（○）（○） （○）（△）

Rules, triangles?
The sum of the internal angles of a triangle is 180 degrees, and the sum of the internal angles of a quadrangle is 360 degrees?

180 * (n-2) n is the number of edges

How to make two squares and three triangles with seven matchsticks

The word "he" should be changed to "or"
If you put two squares together, you can save one and only seven; if you put three regular triangles together, you can save two and also seven

It is known that Sn = n (n + 1) of sequence an, and the nth term BN of sequence BN is equal to the 3N ^ 2 term of sequence an, that is, BN = A3 ^ n
1. Finding the general term an (process) of sequence an
, 2 \ \ find Sn (process) of sequence BN
bn=a3^n

1) An=Sn-S(n-1)=n(n+1)-(n-1)n=2n
2) Bn=A(3^n)=2*3^n
{BN} is an equal ratio sequence with the first term of 6 and the common ratio of 3
Sn=6(1-3^n)/(1-3)=3(3^n-1)

Given an = n / 3N, what is the sum of Sn and the first n terms of the sequence {an}?

Sn=1/3+2/(3^2)+3/(3^3)+.+n/(3^n)
(1/3)Sn=1/(3^2)+2/(3^3)+.+(n-1)/(3^n)+n/[3^(n+1)]
By subtracting the two formulas, we get (2 / 3) Sn = 1 / 3 + 1 / (3 ^ 2) +. + 1 / (3 ^ n) - N / [3 ^ (n + 1)]
=(1/3)[1-(1/3)^n]/[1-(1/3)]-n/[3^(n+1)]
= 1/2-1/2*[(1/3)^n]-n/[3^(n+1)]
Sn=3/4-3/4*[(1/3)^n]-3/2*n/[3^(n+1)]

If Sn = 3n-2, then A3=

a3 = Sn - Sn-1 = 3n-2 - [3(n-1) -2] = 3

"The sum of the first n terms of arithmetic sequence" in Senior One: D = - 1 / 6, Sn = - 5, A1 = 5 / 6, find n and an? Fast! Thank you!

Sn = (a1 + an) * n / 2; = (a1 + A1 + (n-1) d) * n / 2; = A1 * n + (n-1) * n * D / 2; = 5 / 6 * n + (n-1) * n * (- 1 / 6) / 2 = - 5 * 2 * n - (n-1) * n = - 5 * 12 N ^ 2-11n-60 = 0; (N-15) (n + 4) = 0; n = 15, rounding off (n = - 4) n > 0; an = a1 + 14 * d = 5 / 6-14 * (1 / 6) = - 3 / 2;

In the arithmetic sequence {an}, A3 + A6 + A9 = 27, Sn is the sum of the first n terms of the sequence {an}
Then S11 = how much

a3+a9=2a6
a3+a6+a9=27
3a6=27
a6=9
s11=(a1+a11)x11/2=2a6x11/2=11a6=11x9=99

It is known that an is an arithmetic sequence, and A2 = - 8. If the arithmetic sequence BN satisfies B1 = - 8, B2 = a1 + A2 + a3, find the first n terms and TN of BN

b2=a1+a2+a3=3a2=-24
So d = b2-b1 = - 16
bn=-8+(n-1)(-16)=8-16n
Tn=(b1+bn)n/2
=-8n^2

In the arithmetic sequence {an}, it is known that A3 + A8 > 0 and S9
Why is S9 = 9 * A5?

s9=a1+a2+a3+...+a9= 9*a50
∴d>0
∴a4,a3,a2,a10
s5-s4=a5s5
The smallest is S5