Use eight matchsticks to make a small fish. Move three matchsticks to make the fish swim in the opposite direction. How to move

Use eight matchsticks to make a small fish. Move three matchsticks to make the fish swim in the opposite direction. How to move

The picture shows a bird flying to the left with a matchstick. Can you just translate three matchsticks to make it fly to the right? Draw the figure after translation

The picture shows a swallow and a fish made of eight matches
(1) Move the matchstick and change the direction of the swallow and the fish.
(2) If you only move three matchsticks, can you make the swallow fly in the opposite direction and the fish swim in the opposite direction?

See on the net, give you!

As shown in the picture, Xiao Ming uses eight matchsticks to build a small fish, and finds that ∠ 1___ The angle is 2____ Angle

Weak answer: the apposition angle of 1 is the internal angle of 2. I don't know if it's right, I think

As shown in the picture, use matchstick to build small fish in the following way, 8 matchsticks for one small fish, 14 matchsticks for two small fish It is necessary to build n small fish______ A matchstick. (expressed by an algebraic expression containing n)

Number of small fish sticks 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 8 = 8 + 6 (1-1) 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 14 = 8 + 6 (2-1) N & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 8 + 6 (n-1) = 6N + 2, so it takes 8 + 6 (n-1) = 6 to build n small fish

Xiao Ming made a swallow flying pattern with 12 matches. Can you move only 3 matches and make the swallow fly in the opposite direction?
 

Yes

Given that {an} is an equal ratio sequence, A2 = 2, A5 = 14, then A1A2 + a2a3 + +anan+1=（　　）
A. 16（1-4-n）B. 16（1-2-n）C. 323（1-4-n）D. 323（1-2-n）

From A5 = 14 = A2 · Q3 = 2 · Q3, the solution is q = 12. The sequence {Anan + 1} is still an equal ratio sequence: the first term is A1A2 = 8, the common ratio is 14, so A1A2 + a2a3 + +Anan + 1 = 8 [1 - (14) n] 1-14 = 323 (1-4-n), so C

Given that {an} is an equal ratio sequence, A2 = 2, A5 = 14, then A1A2 + a2a3 + +anan+1=（　　）
A. 16（1-4-n）B. 16（1-2-n）C. 323（1-4-n）D. 323（1-2-n）

From A5 = 14 = A2 · Q3 = 2 · Q3, the solution is q = 12. The sequence {Anan + 1} is still an equal ratio sequence: the first term is A1A2 = 8, the common ratio is 14, so A1A2 + a2a3 + +Anan + 1 = 8 [1 - (14) n] 1-14 = 323 (1-4-n), so C

It is known that A1 = 1, an + 1 = 1 + A1A2. An. To prove: (1 / A1) + (1 / A2) +. + (1 / an) = 2

When n tends to infinity, (1 / A1) + (1 / A2) +. + (1 / an) = 2
Just verify that 1 / an = 1 / A1A2... A (n-1) - 1 / A1A2... An
That's fine

It is known that {an} is an equal ratio sequence, an > 0, Sn = a1 + A2 +. An, TN = 1 / A1 + 1 / A2 +. 1 / an, and it is proved that A1A2. An = (Sn / TN) ^ n / 2

Sn=a1(1-q^n)/(1-q)
Tn=1（1-1/q^n)/a1（1-1/q）
a1a2…… an=a1^nq^(1+2+…… +n-1)={a1q^[(n-1)/2]}^n
(sn/Tn)^n/2=[a1^2q^(n-1)]^n/2={a1q^[(n-1)/2]}^n=a1a2…… an