H****h
发帖数: 1037 | 1 1^3+2^3+...+n^3=(1+2+...+n)^2 |
l*****e
发帖数: 238 | 2 我觉得就是碰巧//esc...
【在 H****h 的大作中提到】
: 1^3+2^3+...+n^3=(1+2+...+n)^2
|
s***n
发帖数: 9499 | 3
An geometric interpretation:
Let A(n) = 1+2+...+n
This is a triangle.
Put two triangle togethor, we get a square, thus
n^2 = A(n) + A(n-1).
"Thick" this 2D square to a 3D cubic
n^3 = {A(n) + A(n-1)}*n
= {A(n) + A(n-1)}*{A(n) - A(n-1)}
= A(n)^2 - A(n-1)^2
Add all the cubics togethor.
1^3+2^3+...+n^3 = A(n)^2.
【在 H****h 的大作中提到】
: 1^3+2^3+...+n^3=(1+2+...+n)^2
|
S******g
发帖数: 365 | 4 cool! haha
【在 s***n 的大作中提到】
:
: An geometric interpretation:
: Let A(n) = 1+2+...+n
: This is a triangle.
: Put two triangle togethor, we get a square, thus
: n^2 = A(n) + A(n-1).
: "Thick" this 2D square to a 3D cubic
: n^3 = {A(n) + A(n-1)}*n
: = {A(n) + A(n-1)}*{A(n) - A(n-1)}
: = A(n)^2 - A(n-1)^2
|
l*****e
发帖数: 238 | 5
this does not have any geometric interpretation during construction...
【在 s***n 的大作中提到】
:
: An geometric interpretation:
: Let A(n) = 1+2+...+n
: This is a triangle.
: Put two triangle togethor, we get a square, thus
: n^2 = A(n) + A(n-1).
: "Thick" this 2D square to a 3D cubic
: n^3 = {A(n) + A(n-1)}*n
: = {A(n) + A(n-1)}*{A(n) - A(n-1)}
: = A(n)^2 - A(n-1)^2
|
p*******k
发帖数: 57 | 6
the volume sum of cubics
= the volume of area(=sum of 1 to n)* height(sum of 1 to n).
【在 l*****e 的大作中提到】
:
: this does not have any geometric interpretation during construction...
|
c*******h
发帖数: 1096 | 7 Not quite sure what you mean by triangle. Here is how I interpret the
equality:
Starting at n = 1. Just one dot.
*
When n = 2, expand the dot in two directions, making it a (1+2)*(1+2)
square.
@@@
@@@
*@@
When n = 3, expand further. Make a (1+2+3)*(1+2+3) square.
%%%%%%
%%%%%%
%%%%%%
@@@%%%
@@@%%%
*@@%%%
Each time when you expand the square, you just add something. It turns
out that what you add is n^3. So comes the equality 1^3 + 2^3 + ... +
n^3 = (1+2+...+n)^2.
To see why, think of the example
【在 s***n 的大作中提到】
:
: An geometric interpretation:
: Let A(n) = 1+2+...+n
: This is a triangle.
: Put two triangle togethor, we get a square, thus
: n^2 = A(n) + A(n-1).
: "Thick" this 2D square to a 3D cubic
: n^3 = {A(n) + A(n-1)}*n
: = {A(n) + A(n-1)}*{A(n) - A(n-1)}
: = A(n)^2 - A(n-1)^2
|
m***a
发帖数: 38 | 8 nice
it's just (1+2+...+k)^2 - (1+2+...(k-1))^2 = k^3
then use induction
【在 c*******h 的大作中提到】
: Not quite sure what you mean by triangle. Here is how I interpret the
: equality:
: Starting at n = 1. Just one dot.
: *
: When n = 2, expand the dot in two directions, making it a (1+2)*(1+2)
: square.
: @@@
: @@@
: *@@
: When n = 3, expand further. Make a (1+2+3)*(1+2+3) square.
|
D**u
发帖数: 204 | 9 Here is an ad hoc one:
n+1 couples are in a party, and they want to play a game which needs 2 guys
and 2 ladies. How many different combinations do they have?
One way to count is to compute directly, and the anwser is ((n+1)*n/2)^2,
which is (1+2+...+n)^2.
Another way to count is to see, if a couple (say, Mr. and Mrs. Robinson)
desides not to join the game, then how many combinations are removed?
It is n*n + 2 * n * (n*(n-1)/2) = n^3. Using induction, then the total is
1^3+2^3+...+n^3.
【在 H****h 的大作中提到】
: 1^3+2^3+...+n^3=(1+2+...+n)^2
|
D**u
发帖数: 204 | 10 The equality is equivalent to
(1+2+...+n)^2 - (1+2+...+(n-1))^2 = n^3 = n^2 * n.
We will give an explanation on this formula.
Let X = 1+2+...+n and Y = 1+2+...+(n-1),
then X + Y = n^2, and X - Y = n.
So X + Y = (X - Y) ^2. (1)
Re-arranging (1), we get
X(X-1)/2 + Y(Y-1)/2 = X*Y (2)
The meaning of (2) is:
If there are X boys and Y girls, and we want to pick up 2 of them to form a
2-person team, then there are exactly same number of same-gender teams and
girl-boy teams.
【在 H****h 的大作中提到】
: 1^3+2^3+...+n^3=(1+2+...+n)^2
|
v*******e
发帖数: 3714 | 11 pretty nice. niu.
I was thinking of a directed graph whose vertices are edges
of a complete graph with n+1 vertices..
guys
【在 D**u 的大作中提到】
: Here is an ad hoc one:
: n+1 couples are in a party, and they want to play a game which needs 2 guys
: and 2 ladies. How many different combinations do they have?
: One way to count is to compute directly, and the anwser is ((n+1)*n/2)^2,
: which is (1+2+...+n)^2.
: Another way to count is to see, if a couple (say, Mr. and Mrs. Robinson)
: desides not to join the game, then how many combinations are removed?
: It is n*n + 2 * n * (n*(n-1)/2) = n^3. Using induction, then the total is
: 1^3+2^3+...+n^3.
|
