With six matches placed into a fish, how to move two of the matches, let people see that the fish was eaten by the kitten Represented by graph

Move the top one to the other end, and then move the other one to the middle!

Use eight matches to make two 8-shaped diamonds. How to move two matches to make a diamond?

Put the two matches together, move the two matches at the lower part of the left one, move them into the words "one" and "one" (don't say you can't do either), and then combine them together to form the pattern of "one"
They said it

First, use 23 matches to make it look like 2000, and then please move 8 matches to make it a province of our country. How to move it

Shanxi

How to make 4 triangles with 7 matches

I only need six, please see the answer. Take three to make an equilateral triangle, and then take the fourth to put the middle point on the top of the equilateral triangle, and make it parallel to the bottom of the equilateral triangle. Take the fifth and sixth matches, connect the end of the fourth match with the middle point of the bottom of the equilateral triangle, and you can get four small equilateral triangles

A1A2. An belongs to positive real number. Given a1 + A2 +. An = 1, prove A1'2 / A1 + A2 +. An2 / an + an1 ≥ 1 / 2

Using the generalized Cauchy inequality
a1^2/(a1+a2) +.+an^2/(an+an1)
>=(a1+a2+…… +an)^2/(2a1+2a2+…… +2 an) = 1 / 2

An = 4 ^ n-3 ^ n prove 1 / A1 + 1 / A2 + 1 / A3 + 1/an
Expert help an = 4 ^ n-3 ^ n prove 1 / A1 + 1 / A2 + 1 / A3 + 1/an

It is suggested that the method of expansion and contraction is generally used in this kind of problems. The proof: [1 / a (n + 1)] / (1 / an) = an / a (n + 1) = (4 & # 8319; - 3 & # 8319;) / [4 ^ (n + 1) - 3 ^ (n + 1)] = (1 / 3) (3 × 4 & # 8319; - 3 × 3 & # 8319;) / [4 ^ (n + 1) - 3 ^ (n + 1)] = (1 / 3) [4 ^ (n + 1) - 3 ^ (n + 1) - 4 & # 8319;] / [4 ^ (n + 1) - 3 ^ (n + 1)] = (1 / 3) = (4 ^ (n + 1) = (1) / [4 ^ (n + 1) - 3 ^ (n + 1)] =

The general formula of the sequence {an} {BN} is an = n-16, BN = (- 1) n power · ｜ N-15 ｜, where n ∈ n *
Q: remember that the sum of the first n terms of the sequence {anbn} is Sn, and find all ordered integer pairs (m, n) satisfying S2M = s2n (m < n)

anbn=(n-16) *(-1)^n *ln-15l
When n > 16, anbn = (- 1) ^ n * (n-16) (N-15) = (- 1) ^ n * (n ^ 2-31n + 16 * 15)
Anbn = 0 when n = 15,16
When n

The sequence an, BN satisfies BN = a1 + 2A2 + 3a3... Nan + 1 + 2 + 3 +... N. if BN is an arithmetic sequence, it is proved that an is an arithmetic sequence

Certification:
First, we simplify the formula: a1 + 2A2 + 3a3... + Nan = BN * (1 + 2 + 3 +... + n) = BN * n (n + 1) / 2
Take n-1 term, so a1 + 2A2 + 3a3... + (n-1) a (n-1) = B (n-1) * n (n-1) / 2
The corresponding left-right subtraction of the two formulas is: Nan = BN * n (n + 1) / 2-B (n-1) * n (n-1) / 2
Divide both sides by n to get an = BN * (n + 1) / 2-B (n-1) * (n-1) / 2 = [(n + 1) BN - (n-1) B (n-1)] / 2
Suppose that BN is an arithmetic sequence, let bn-b (n-1) = D (constant),
So an = [Nd + BN + B (n-1)] / 2
So an-a (n-1) = 3D / 2, that is, an is an arithmetic sequence

The sequence {an} {BN} satisfies the relation BN = 1 * a1 + 2 * A2 + 3 * A3 +nan/1+2+3+… +n. If {BN} is an arithmetic sequence, prove that {an} is also an arithmetic sequence

Proof: let {BN} tolerance be d (D is a constant). A1 / 1 = B1, A1 = b1bn = (a1 + 2A2 + 3a3 +... + Nan) / (1 + 2 + 3 +... + n) a1 + 2A2 + 3a3 +... + Nan = [n (n + 1) / 2] BN (1) a1 + 2A2 + 3a3 +... + (n + 1) a (n + 1) = [(n + 1) (n + 2) / 2] B (n + 1) (2) (2) (2) - (1) (n + 1) a (n + 1) = [(n + 1) (n + 2) / 2] B (n + 1) = [(n + 1) (n + 2) / 2]

bn=a1+2a2+3a3+4a4+…… +Nan if an is an arithmetic sequence, then BN =?

The sequence {an} is a positive arithmetic sequence if BN = (a1 + 2A2 + 3a3 +...) +nan)/(1＋2＋3＋… +n) Then the sequence {BN} is also an arithmetic sequence;
Let an tolerance be D, then
bn=(a1+2a2+3a3+… +nan)/(1+2+3+… +n)
=2(a1+2a2+3a3+… +nan)/n(n+1)
=2(a1+2(a1+d)+3(a1+2d)+… +n(a1+(n-1)d)/n(n+1)
=2{(a1+2a1+3a1+… +na1)+[1*2+2*3+3*4+… (n-1)n]d}/n(n+1)
=2{(n(n+1)a1/2)+[1*2+2*3+3*4+… (n-1)n]d}/n(n+1)
={(n(n+1)a1)+2[1*2+2*3+3*4+… (n-1)n]d}/n(n+1)
=a1+2[1*2+2*3+3*4+… +(n-1)n]d/n(n+1)
=a1+2[1+2+3+… +n-1+1^2+2^2+3^2+… +(n-1)^2]d/n(n+1)
=a1+2(n-1)n(n+1)d/3n(n+1)
=a1+(n-1)2d/3
That is, BN is an arithmetic sequence with A1 as the first number and 2D / 3 as the tolerance
bn＝a1+2a2+3a3+… nan/1+2+3… +n
b(n+1)=[a1+2a2+3a3+… nan+(n+1)a(n+1)]/[1+2+3… +n+(n+1)]
[n(n+1)/2]bn=a1+2a2+3a3+… nan ①
[(n+1)(n+2)/2]b(n+1)=a1+2a2+3a3+… nan+(n+1)a(n+1) ②
② - 1
[(n+1)(n+2)/2]b(n+1)-[n(n+1)/2]bn=（n+1)a(n+1)
If (n + 1) is deleted on both sides at the same time, then
a(n+1)=[(n+2)/2]b(n+1）-(n/2)bn③
an=[(n+1)/2]bn-[(n-1)/2]b(n-1) ④
③ (4) a (n + 1) - an = [(n + 1) / 2] B (n + 1) + 1 / 2B (n + 1) - [(n + 1) / 2] BN - [(n-1) / 2] BN + [(n-1) / 2] B (n-1) - 1 / 2bn
=[(n+1)/2][b(n+1)-bn]+1/2[b(n+1）-bn]-[(n-1)/2][bn-b(n-1)]
And {BN} is the arithmetic sequence, and the tolerance is d
Then a (n + 1) - an = [(n + 1) / 2] d + 1 / 2 * D - [(n-1) / 2] D
=3/2d
So {an} is an arithmetic sequence with tolerance of 3 / 2D
Note: an, BN, a (n + 1), B (n + 1) are all terms in the sequence

With six matches placed into a fish, how to move two of the matches, let people see that the fish was eaten by the kitten Represented by graph