"Counting in their heads" - 1895 oil painting
08-14-2020, 02:25 PM (This post was last modified: 08-14-2020 02:29 PM by Albert Chan.)
Post: #23
 Albert Chan Senior Member Posts: 2,231 Joined: Jul 2018
RE: "Counting in their heads" - 1895 oil painting
(08-12-2020 01:32 AM)Albert Chan Wrote:  I was wrong. There seems to be a pattern to sum of powers formula after all ...

$$S_p = n c^p + \large {n^3-n \over 12} \left( \binom{p}{2} c^{p-2} + {3n^2-7 \over 20} \binom{p}{4} c^{p-4} + {3n^4-18n^2+31 \over 112} \binom{p}{6} c^{p-6} + {5n^6-55n^4+239n^2-381 \over 960} \binom{p}{8} c^{p-8} + \cdots \right)$$

Example:

50^5 + 51^5 + 52^5 + ... + 150^5 ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// p=5, c=100, n=101

= 101*100^5 + 102*101*100/12 * (10*100^3 + (3*101^2-7)/20*5*100)
= 1934166665000

To extend formula for spacings = d, simply replace Sp by Sp/dp, c by c/d
Or, just check the dimensions for each term. All terms must have same units.

Example, for sum-of-squares, all terms should have unit of c²

S2 = n*c² + (n³-n)/12 * ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ // c/d is dimensionless, thus c, d have same unit

Example, for sum of m odd squares

1² + 3² + 5² + ... + (2m-1)² ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// d=2, c = m = n
= m*m² + m*(m²-1)/12 * 2²
= m*(4m²-1)/3
= $$\binom{2m+1}{3}$$

We can confirm this from sum-of-n-squares formula

\begin{align} {n(n+1)(2n+1) \over 6} &= {n(n+1)·[(n-1) + (n+2)] \over 6}\\ \binom{2n+2}{3}/4 &= \binom{n+1}{3} + \binom{n+2}{3}\\ \end{align}

Let n = 2m:
LHS = sum-of-m-odd-squares + sum-of-m-even-squares
$$\binom{n+2}{3}$$ = 4×sum-of-m-squares = sum-of-m-even-squares

⇒ $$\binom{2m+1}{3}$$ = sum-of-m-odd-squares
 « Next Oldest | Next Newest »