A New Algebraic Relation Between Sequential Natural Number Sums

Posted on: March 9, 2025

We know how to calculate the sum of the first natural numbers. If our sequence doesn't start with 1, then we can use the arithmetic series formula to find any series' nth sum.

By using these two formulas, we can find any even-length series sum with increment value d = 1 by just calculating its first half sum. Let's see how we can do this.

Consider any series with r = 1.
Let's take this: 7, 8, 9, 10, 11, 12, where N = 6.
Let a be the first half sum of this series, i.e., 7+8+9.
Let b be the second half sum of this series, i.e., 10+11+12.
We can find b by just adding (N/2)² to the value a:

\[ b = a + (N/2)^2 \]

Let's say n = N/2, then:

\[ b = a + n^2 \]

The sum of the series would be:

\[ T = a + b = 2a + n^2 \]

For example: T = 2(24) + 3² = 48 + 9 = 57

We're not stopping here. Let's do the same for the square series as well.

Consider any series with r = 1.
Let's take this: 7², 8², 9², 10², 11², 12², where N = 6.
Let x be the first half sum of this series, i.e., 7² + 8² + 9².
Let y be the second half sum of this series, i.e., 10² + 11² + 12².
Then:

\[ y = x + nT \]

Where T is the sum of the original sequence: 7+8+9+10+11+12.
The sum of the squared series will be:

\[ T_2 = x + y = 2x + nT \]

Let's expand this formula to make the calculation even easier:

\[ T = \frac{N}{2}(2A + (N-1)d) \text{ (arithmetic series formula)} \]

Where A is the starting term, N = 2n, and d = 1.

\begin{align} T_2 &= 2x + n\left(\frac{N}{2}(2A + N - 1)\right)\\ &= 2x + n(n(2A + 2n - 1))\\ &= 2x + 2An^2 + 2n^3 - n^2\\ &= 2(x + n^3) + n^2(2A-1) \end{align}

Example:

x = 7²+8²+9² = 194
A = 7
n = 3

\begin{align} T_2 &= 2(194 + 27) + 9(13)\\ &= 2(221) + 117\\ &= 442 + 117\\ &= 559 \end{align}

= 7²+8²+9²+10²+11²+12²

Let's derive a formula for the first n natural numbers:

For the first n natural numbers, A will be 1

\[ S_n = 2a + n^2 \quad \text{(from our first formula)} \]

Applying a value, we get:

\[ S_n = 2n^2 + n \]

So:

\[ a + b = 2n^2 + n \]

But we also know:

\[ An^2 = \sqrt{b-a} \]

Since A = 1:

\begin{align} a + b &= 2(b-a)+\sqrt{b-a}\\ 3a - b &= \sqrt{b-a} \end{align}

Squaring both sides:

\begin{align} 9a^2+b^2-2(3a)(b) &= b-a \end{align}

After simplification:

\[ \frac{9a+1}{6b} + \frac{b-1}{6a} = 1 \]

This equation holds true for all natural number series starting with 1.