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.
