Solving the Sum ( sum_{k1}^{n} sum_{m1}^{k} frac{m(m-1)}{2} ) Using Polynomial Methods

The given sum ( sum_{k1}^{n} sum_{m1}^{k} frac{m(m-1)}{2} ) involves a quadratic term and can be represented as a fourth-degree polynomial in ( n ). In this article, we will explore the solution using both algebraic and combinatorial methods.

Algebraic Solution

To solve the given sum using polynomial methods, we will use the properties of polynomials and linear algebra. When summing a quadratic function, the result is a cubic function, and when summing a cubic function, the result is a quartic function. Hence, we can write the sum as a fourth-degree polynomial in ( n ), which we denote as:

[ f(n) a n^4 b n^3 c n^2 d n ]

We can determine the coefficients ( a ), ( b ), ( c ), and ( d ) by evaluating ( f(n) ) at specific values of ( n ).

Evaluating the Coefficients

We start by evaluating ( f(1) ), ( f(2) ), ( f(3) ), and ( f(4) ) to form a linear system of equations:

[ f(1) 1 ]

[ f(2) 5 ]

[ f(3) 15 ]

[ f(4) 35 ]

We can write these values in terms of the coefficients:

[ a b c d 1 ]

[ 16a 8b 4c 2d 5 ]

[ 81a 27b 9c 3d 15 ]

[ 256a 64b 16c 4d 35 ]

Solving the Linear System

We can solve this system using linear algebra techniques. Here is a line of MATLAB/Octave code to find the coefficients:

```matlab coeffs rats(inv(bsxfun(@power, 1:4, [1 5 15 35])); fliplr(1:4)) ```

The result is:

[ coeffs begin{cases} frac{1}{24} frac{1}{4} frac{11}{24} frac{1}{4} end{cases} ]

Therefore, the polynomial can be expressed as:

[ f(n) frac{1}{24} n^4 frac{1}{4} n^3 frac{11}{24} n^2 frac{1}{4} n ]

After some factoring, we can simplify this to:

[ f(n) frac{n(n-1)(n-2)(n-3)}{24} ]

Combinatorial Solution

The combinatorial solution provides a more intuitive understanding of the sum by interpreting it in a combinatorial context.

Interpreting the Sum

The given expression can be broken down as:

[ sum_{k1}^{n} sum_{m1}^{k} frac{m(m-1)}{2} sum_{k1}^{n} sum_{m1}^{k} sum_{p1}^{m} p ]

We then recognize the expression ( sum_{k1}^{n} sum_{m1}^{k} sum_{p1}^{m} p ) as the number of ways to choose four balls from ( n 3 ) balls (where the balls are labeled from 1 to ( n 3 )) and then subtracting the largest number first.

The combinatorial interpretation gives us:

[ binom{n 3}{4} frac{n(n-1)(n-2)(n-3)}{24} ]

Conclusion

Both the algebraic and combinatorial methods provide the same result:

[ sum_{k1}^{n} sum_{m1}^{k} frac{m(m-1)}{2} frac{n(n-1)(n-2)(n-3)}{24} ]

The method we used here is quite generic and can be applied to find explicit expressions for other series involving polynomial sums. Such techniques are particularly useful in combinatorics and can provide elegant solutions to complex summations.