Determinar em função de n a soma das entradas da tabela triangular 

\left(a_{ij}\right) _{n\geq j\geq i\geq 0},

 em que a_{ij}=j.

Resolução

Esta tabela tem a configuração seguinte

\begin{array}{cccccc}0&1&\cdots&j&\cdots&n\\&1&\cdots&j&\cdots&n\\&&\ddots&\vdots&\cdots&\vdots\\&&&j& \cdots&n\\&&&&\ddots&\vdots\\&&&&&n\end{array}

Aplicando os resultados de Somatórios com várias parcelas e as propriedades do somatório duplo, vem sucessivamente

\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i}^{n}j=\displaystyle\sum_{i=0}^{n}\left(\displaystyle\sum_{j=1}^{n}j-\displaystyle\sum_{k=1}^{i-1}k\right)=\displaystyle\sum_{i=0}^{n}\frac{1}{2}\left[n\left( n+1\right) -\left( i-1\right) i\right]

=\displaystyle\frac{1}{2}\left[ n\left( n+1\right)\displaystyle\sum_{i=0}^{n}1-\displaystyle\sum_{i=1}^{n}i^{2}+\displaystyle\sum_{i=1}^{n}i\right]

=\displaystyle\frac{1}{2}\left[ n\left( n+1\right) \left( n+1\right) -\displaystyle\frac{n\left(n+1\right)\left(2n+1\right) }{6}+\displaystyle\frac{n\left( n+1\right) }{2}\right]

=\displaystyle\frac{1}{2}n\left( n+1\right) \left[ \left( n+1\right) -\displaystyle\frac{\left(2n+1\right) }{6}+\frac{1}{2}\right]

=\displaystyle\frac{n\left( n+1\right) \left( n+2\right) }{3}+\displaystyle\frac{n\left( n+1\right) }{2}

=\displaystyle\frac{n\left( n+1\right) \displaystyle\left( n+2\right) }{3}. \qquad\blacktriangleleft