r***e
发帖数: 31 | 1 How do we get the formula:
1^2+2^2+3^3+....+n^2=n(n+1)(2n+1)/6.
I don't mean proof, I mean how to get this formula?
and also for
1^3+2^3+3^3+.....n^3?
forgot a lot of things since I learned it long time ago. |
S******g
发帖数: 365 | 2 最简单的方法，对(n+1)^k展开，然后全加起来，可以在已知1\cdots (k-2)情况下求出
(k-1)的情况。
【在 r***e 的大作中提到】
: How do we get the formula:
: 1^2+2^2+3^3+....+n^2=n(n+1)(2n+1)/6.
: I don't mean proof, I mean how to get this formula?
: and also for
: 1^3+2^3+3^3+.....n^3?
: forgot a lot of things since I learned it long time ago.
|
w***s
发帖数: 1026 | 3 观察找规律：）
【在 r***e 的大作中提到】
: How do we get the formula:
: 1^2+2^2+3^3+....+n^2=n(n+1)(2n+1)/6.
: I don't mean proof, I mean how to get this formula?
: and also for
: 1^3+2^3+3^3+.....n^3?
: forgot a lot of things since I learned it long time ago.
|
b**g
发帖数: 335 | 4 Easy.
Let F(n) = (n+1)-n = 1
G(n) = (n+1)^2- n^2 = 2n+1
H(n) = (n+1)^3 - n^3 = 3n^2+3n+1
Easy to see
3n^2 = H(n) - 3G(n)/2 + F(n)/2
Note that H(1)+...+H(n) is easy to compute (telescoping sum)
and so are G(1)+...+G(n) and F(1)+...+F(n)
【在 r***e 的大作中提到】
: How do we get the formula:
: 1^2+2^2+3^3+....+n^2=n(n+1)(2n+1)/6.
: I don't mean proof, I mean how to get this formula?
: and also for
: 1^3+2^3+3^3+.....n^3?
: forgot a lot of things since I learned it long time ago.
|
w***s
发帖数: 1026 | 5 niubi
【在 b**g 的大作中提到】
: Easy.
: Let F(n) = (n+1)-n = 1
: G(n) = (n+1)^2- n^2 = 2n+1
: H(n) = (n+1)^3 - n^3 = 3n^2+3n+1
: Easy to see
: 3n^2 = H(n) - 3G(n)/2 + F(n)/2
: Note that H(1)+...+H(n) is easy to compute (telescoping sum)
: and so are G(1)+...+G(n) and F(1)+...+F(n)
:
|
H****h
发帖数: 1037 | 6 知道一定是多项式。然后凑系数。
【在 b**g 的大作中提到】
: Easy.
: Let F(n) = (n+1)-n = 1
: G(n) = (n+1)^2- n^2 = 2n+1
: H(n) = (n+1)^3 - n^3 = 3n^2+3n+1
: Easy to see
: 3n^2 = H(n) - 3G(n)/2 + F(n)/2
: Note that H(1)+...+H(n) is easy to compute (telescoping sum)
: and so are G(1)+...+G(n) and F(1)+...+F(n)
:
|
w***s
发帖数: 1026 | 7 degree呢？
【在 H****h 的大作中提到】
: 知道一定是多项式。然后凑系数。
|
r***e
发帖数: 31 | 8 cool, thx for the answer. |
H****h
发帖数: 1037 | 9 不超过k+1。
【在 w***s 的大作中提到】
: degree呢？
|
s***n
发帖数: 9499 | 10
Triangle
1 + 2 + ... + n = C(n+1, 2)
Which is the combination of choose 2 from n+1.
Think geometrically, you can decompose the
1^2+2^2+3^3+....+n^2 to two tetrahedrals, thus
1^2+2^2+3^3+....+n^2 = C(n+2, 3) + C(n+1, 3)
=n(n+1)(2n+1)/6.
done
【在 r***e 的大作中提到】
: How do we get the formula:
: 1^2+2^2+3^3+....+n^2=n(n+1)(2n+1)/6.
: I don't mean proof, I mean how to get this formula?
: and also for
: 1^3+2^3+3^3+.....n^3?
: forgot a lot of things since I learned it long time ago.
|
