## Two exceedingly simple sums related to triangular numbers

This note records some elementary arithmetic pertaining to triangular numbers for bālabodhana. In our youth we found that having a flexible attitude was good thing while obtaining closed forms for simple sums: for some sums geometry (using methods of proofs pioneered by Āryabhaṭa which continued down to Nīlakaṇṭha Somayājin) was the best way to go; for others algebra was better. The intuition was in choosing the right approach for a given sum. We illustrate that with two such sums.

Sum 1 Obtain a closed form for the sum: $\displaystyle \sum_{j=1}^{n} (2j-1)^3$

These sums define a sequence: 1, 28, 153, 496, 1225…
Given that we can mostly only visually operate in 3 spatial dimensions, our intuition suggested that a cubic sum as this is best tackled with brute-force algebra with the formulae for individual terms derived by Āryabhaṭa and his commentators. Thus we have:

$\displaystyle \sum_{j=1}^{n} (2j-1)^3 = \sum_{j=1}^{n} 8 j^3 - 12 j^2 + 6 j- 1$
$= 2n^2(n+1)^2-2n(n+1)(2n+1)+3n(n+1)-n= \dfrac{(2n^2-1)2n^2}{2}$

The reason we wrote out the final solution in this unsimplified form is to illustrate that the above sums will always be a triangular number of the form:

$\displaystyle \sum_{j=1}^{2j^2-1} j$, i.e sums from 1 to 1, 7, 17, 31, 49… or triangular numbers $T_1, T_7, T_{17}, T_{31}\cdots$

Thus, the $n$th terms of sequence of sums would be triangular number $T_m$, where $m=2j^2-1, j=1, 2, 3...$. From the above, one can also see that the difference of successive terms of our original sequence of sums will be 27, 125, 343, 729…, i.e., they are perfect cubes of the form $(2k+1)^3$ (odd numbers 3, 5, 7, 9…). These cubes are thus the interstitial sums of the indices $j$ of the triangular numbers $T_j$ up to the index $m$ corresponding to the triangular number $T_m$ that is a term of our original sequence. Thus:
$j\mapsto$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31
then, 2+3+4…+7=27; 8+9+10…+17=125; 18+19+20+21+22…+31 =343 and so on.

Another interesting feature of the original sequence is its decadal cycle in the terms of the numbers in the last 2 places (written in anti-Hindu, i.e. modern order). They will always end in the following sequence of 10 numbers:
0, 1, 28, 53, 6, 25, 6, 53, 28, 1
Similarly, the index $m$ of the triangular numbers $T_m$ the define our sequence also shows a pentadic cycle in the last place of the form:
1, 7, 7, 1, 9

A comparable pattern is seen if we generate a sequence that is the sum of successive terms of our original sequence: 1, 29, 181, 649, 1721, 3781, 7309, 12881… The last place has a pentadic cycle of the form: 1,9,1,9,1. The last 2 places has a cycle of length 25: 01, 29, 81, 49, 21, 81, 09, 81, 69, 41, 61, 89, 81, 89, 61, 41, 69, 81, 09, 81, 21, 49, 81, 29, 01. Both are palindromic cycles.

Finally, the sum of the reciprocals of the original sequence converges to a constant: 1.04607799646… We suspect there is a closed form for this constant but have not been able to identify it.

Sum 2 Obtain a closed form for the sum of alternating negative and positive perfect squares: -1+4-9+16… i.e.

$\displaystyle \sum_{j=1}^n (-1)^j j^2$

With the sum involving just square terms it is possible to use a wordless geometric proof along the lines of that proposed by Āryabhaṭa (Figure 1).

Figure 1.

Thus, we get the above sum as $-1^n T_n$, where $T_n$ is the $n$th triangular number.

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