Previously, we saw the मूलसङ्कलितम् \( S_n = 1+2+…n=\frac{n.(n+1)}{2} \) and also noted that as *n *becomes large, \(S_n \approx \frac{n^2}{2}\). We also saw the समघातसङ्कलितानि, sums of higher powers, including the sum of squares – वर्गसङ्कलितम्, and sum of cubes – घनसङ्कलितम्. We noted that in the limit of large *n*, the kth power sankalita tends to:

S_n^{(k)} \approx \frac{n^{k+1}}{k+1}

\)

We also saw सङ्कलितसङ्कलितम्, the second order sankalita which is the sum of first order sankalitas \(

SS^{(2)}_n = S_n + S_{n-1} + …. + 1

\), and also other higher order sankalitas.

### Summary of results

For large n, Yuktibhāṣā proves the following results for सङ्कलितानि:

\(

\begin{align*}

S_n &= 1+2+3+…+n \\

&\approx \frac{n^2}{2} \\

S^{(2)}_n &= 1^2+2^2+3^2+…n^2 \\

&\approx \frac{n^3}{3} \\

… \\

S^{(k)}_n &= 1^k+2^k+3^k+…n^k \\

&\approx \frac{n^{k+1}}{k+1} \

\end{align*}

\)

and also for सङ्कलितसङ्कलितानि:

\( \begin{align*}

SS^{(2)}_n &\approx \frac{n^3}{6}\\

SS^{(3)}_n &\approx \frac{n^4}{24}\\

…\\

SS^{(k)}_n &\approx \frac{n^{k+1}}{(k+1)!}\

\end{align*} \)

## The Triangle Law of Sankalitas

To help us with the proof of Madhava's sankalita for jyā, we will note an interesting, but simple result for sankalitas, which I have taken the liberty of calling the Triangle Law

\(

\begin{align*} SS^{(k)}_n &= SS^{(k-1)}_n + SS^{(k-1)}_{n-1} + …. + 1 \\

&= SS^{(k-2)}_n + SS^{(k-2)}_{(n-1)} + SS^{(k-2)}_{(n-2)} + … + 1 \\

&\quad+ SS^{(k-2)}_{(n-1)} + SS^{(k-2)}_{(n-2)} + … + 1 \\

&\quad+ SS^{(k-2)}_{(n-2)} + … + 1 \\

&\quad … \\

&\quad + 1 \\

& = SS^{(k-2)}_n + 2.SS^{(k-2)}_{(n-1)} + 3.SS^{(k-2)}_{(n-2)} + … n.1 \\

& = \sum_{j=1}^{n}j.SS^{(k-2)}_{(n-(j-1))}

\end{align*}

\)

This result allows us to express the *kth* sankalita-sankalita as a sum of *(k-2)th *sankalita-sankalitas, in addition to the elementary expression in terms of *(k-1)th *sankalita-sankalitas, which we will find useful.