Mahaviracharya’s Series: Deriving the Sum of Squares in Natural Numbers

As mathematics enthusiasts, we often apply various formulas to solve problems and explore mathematical concepts. One such noteworthy formula is the sum of squares of the first \(n\)  natural numbers:

\[1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\]

While many of us may have encountered and derived this formula through the identity

\[k^3 – (k-1)^3 = 3k^2 -3k + 1 \tag*{—– (1)} \]

I would like to introduce an alternative derivation. In this approach, we will delve into a fascinating series presented by Mahaviracharya, an extension of his original work on the formula for finding the cube of any number, as initially provided by Sridharacharya. This alternative perspective offers a unique insight into the sum of squares of natural numbers.

Sridharacharya’s Formula

According to Sridharacharya, the cube of any number \(n\) is expressed as one more than the sum of the cube of the number immediately preceding \( n \) and three times the product of \( n \) and the preceding number. This relationship is formally represented by the equation:

\[ n^3 = (n-1)^3 + 3n(n-1) + 1 \tag*{—– (2)} \]

For instance, when determining the cube of \( 5 \), the formula is applied as follows:

\[ 5^3 = 4^3 + 3 \times 5 \times 4 + 1 \]

This formula facilitates the connection between the cubes of two consecutive numbers. Notably, equations (1) and (2) are equivalent.

Consequently, if \( n = m+1 \), the formula (2) can be expressed as:

\[ n^3 = m^3 + 3m(m+1) + 1 \tag*{—– (3)} \]

This formulation retains the essential information while presenting it in a more polished and structured manner.

Mahaviracharya’s Exploration

Mahaviracharya, in his profound mathematical exploration, expanded upon this mathematical identity initially proposed by Sridharacharya. His insightful extension involved establishing a profound connection between the cube of a number \(n\) and a sum involving numbers less than a value \(m\). Through his meticulous analysis, he unveiled a remarkable expression, demonstrating that the cube of any given number \(n\) could be elegantly expressed as:

\[n^3 = 3s + n,\]

where \(s\) is defined as the sum:

\[s = n(n-1) + (n-1)(n-2) + \cdots + 3 \times 2 + 2 \times 1.\]

This captivating formula offers a fascinating insight into the cubic relationships of numbers. To illustrate, consider the following examples:

\[5^3 = 3(5 \times 4 + 4 \times 3 + 3 \times 2 + 2 \times 1) + 5,\]

\[3^3 = 3(3 \times 2 + 2 \times 1) + 3,\]

and so forth. Notably, Mahaviracharya’s formulation possesses the remarkable quality of rendering the right-hand side entirely devoid of any cubic expression, elevating the elegance and simplicity of the mathematical representation.

Foundation for \(s\) that is a summation of \(n\) pairs:

The origin of this sum can be illuminated by examining equation (2):

\[n^3 = (n-1)^3 + 3n(n-1) + 1 \tag*{—– (2)}\]

Delving into the intricacies of this equation sheds light on the essence of \(s\).

This formula (2) can be systematically applied to every cubic expression on the right-hand side, unfolding a cascade of interconnected relationships. Each iteration of this formula not only preserves the cubic nature of the expression but also contributes to the construction of the cumulative sum \(s\).

We embark on a captivating journey of recursive expansion, unraveling the layers of cubic expressions. The equation

\[n^3 = (n-1)^3 + 3n(n-1) + 1 \tag*{—– (2)}\]

initiates this mesmerizing pattern. Applying the same logic to the expression \((n-1)^3\) yields

\[(n-1)^3 =  (n-2)^3 + 3(n-1)(n-2) +1 \tag*{—– (3)}\]

similarly, expression \((n-2)^3\) yields

\[(n-2)^3 =  (n-3)^3 + 3(n-2)(n-3) +1 \tag*{—– (4)}\]

From the above equations (2), (3) and (4), we get

\[n^3 =  (n-3)^3 + 3\{n(n-1) + (n-1)(n-2) + (n-2)(n-3) \} + 3\]

In this iterative dance of equations, the unfolding pattern is eloquently consistent. Each step in the sequence echoes the previous one, mirroring the elegant structure inherent in cubic expressions. This gives,

\[n^3 = 3\{n(n-1)+(n-1)(n-2)+ \cdots+3 \times 2 + 2\times 1 \} + n\]

thus,

\[s = n(n-1) + (n-1)(n-2) + \cdots + 3 \times 2 + 2 \times 1.\]

Another way to achieve the Same is as under:

Now, let us proceed to derive the desired formula from this insightful expression, delving deeper into the intricacies of Mahaviracharya’s mathematical revelation.

Sum of square of \(n\) numbers

We have the equation

\[n^3 = 3s + n,  \, \text{where} \, s = n(n-1) + (n-1)(n-2) + \cdots + 3 \times 2 + 2 \times 1\]

Firstly, let us unveil the essence of \(s\). To achieve this, we rewrite the \(s\) as:

\[s = 1 \times 2 + 2 \times 3 + \cdots + n(n-1).\]

In this, the series’ \(k\)-th term manifests as \(a_{k} = k(k-1)\). Consequently, \(s = \sum_{k=1}^{n} k(k-1)\), or equivalently,

\[s = \sum_{k=1}^{n} k^2 – \sum_{k=1}^{n} k.\]

Notably, the second sum on the right side, \(\sum_{k=1}^{n} k = 1 + 2 + \cdots + n = \frac{n(n+1)}{2}\), contributes a striking note to our symphony of mathematical expression.

Thus, the equation \(n^3 = 3s + n\) gracefully transforms into the form:

\[n^3 = 3\left(\sum_{k=1}^{n} k^2 – \frac{n(n+1)}{2}\right) + n.\]

Rearranging for the term containing the summation sign, a  formula begins to emerge:

\[\sum_{k=1}^{n} k^2 = \frac{n^3 – n}{3} + \frac{n(n+1)}{2}.\]

And as the curtains draw open on this mathematical sonnet, the desired formula gracefully takes center stage:

\[\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}.\]

In the captivating narrative of mathematical exploration, we’ve ventured through the rich tapestry of Indian mathematical history, illuminating the graceful interplay of ideas and insights. This formula

\[n^3 = 3\{n(n-1)+(n-1)(n-2)+ \cdots+3 \times 2 + 2\times 1 \} + n\]

is like a shining jewel, showing us the incredible beauty hidden in the world of numbers and reflecting the wisdom of ancient Indian mathematicians.

Stay Mathactive! 🙏