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"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
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
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RE: "Counting in their heads" - 1895 oil painting - Albert Chan - 08-14-2020 02:25 PM



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