How to calculate this kind of summation?

Question

I'm now on the first chapter of the book 'Mathematical Methods For Physics and Engineering' and I came across this summation.

$\displaystyle\sum\limits_{n=1}^{n}$$\displaystyle\sum\limits_{k>j}^{n} a_ja_k$

How to do this computation ?I tried searching for an explanation in the internet but I got none.

Answer 1

Downloaded the book. https://www.zuj.edu.jo/download/mathematical-methods-for-physics-and-engineering-riley-hobson-pdf/

Are you doing equation 1.14?

$\sum\limits_{j=1}^n\sum\limits_{k>j}^n \alpha_j\alpha_k = \frac {a_{n-2}}{a_n}$?

In the case the summation means:

$\sum\limits_{j=1}^n\sum\limits_{k>j}^n \alpha_j\alpha_k=$

$\sum\limits_{j=1}^n(\sum\limits_{k>j}^n \alpha_j\alpha_k)=$

$\sum\limits_{j=1}^n(\color{blue}{\sum\limits_{k\text{ is any natural number} > j\text{ up to the final summand of } n}^n} \alpha_j\alpha_k)=$

$\sum\limits_{j=1}^n(\sum\limits_{k=j+1}^n \alpha_j\alpha_k)=$

$\sum\limits_{j=1}^n(\alpha_j\alpha_{j+1} + \alpha_j\alpha_{j+2} + ..... + \alpha_j\alpha_n) = $

$\color{blue}{\sum\limits_{j\text { is every natural number from } 1\text{ to } n}^n}(\alpha_j\alpha_{j+1} + \alpha_j\alpha_{j+2} + ..... + \alpha_j\alpha_n) = $

$(\alpha_1\alpha_2 + ................... + \alpha_1\alpha_n) + $

$(\alpha_2\alpha_3 + ........... + \alpha_2\alpha_n) + $

.........

$(\alpha_{n-2}\alpha_{n-1} + \alpha_{n-2}\alpha_n) + $

$(\alpha_{n-1}\alpha_{n})$

....

In short this is the way we write "the sum of all possible $\alpha_j\alpha_k$ where $k > j$ and and $j,k \le n$".

Answer 2

$$\sum_{j=1}^n\sum_{k=j+1}^n1=\sum_{j=1}^n(n-(j+1)+1)=\sum_{j=1}^n(j-1)$$