1. ## Double Summation question

For all $n\in{N}$ find a formula for $\sum{\sum{a_{ij}}}
$
(first sum between j=1 and n, second between i=1 and j) in each of the following cases:

i) $a_{ij}$ = 1
ii) $a_{ij}$ = ij

2. all you have to remember is the expression for a sum of an arithmetic progression (or use google):

Here's the second case:

$\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {ij} = } \sum\limits_{j = 1}^n {j\sum\limits_{i = 1}^n i = } \left[ {\left( {1 + n} \right)\frac{n}
{2}} \right]^2$

3. Sorry, I think you misunderstood the limits. My fault for not knowing the code for it.

It's this:

$\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j} {a_{ij}}$

for 1 and ij.

4. $
\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j 1 = } \sum\limits_{j = 1}^n {j = \left( {1 + n} \right)\frac{n}
{2}}
$

$\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j {ij} = \sum\limits_{j = 1}^n {j\left( {1 + j} \right)\frac{j}
{2} = } } \frac{1}
{2}\left[ {\sum\limits_{j = 1}^n {j^2 + } \sum\limits_{j = 1}^n {j^3 } } \right]$

these are well known sums Power Sum -- from Wolfram MathWorld

5. Originally Posted by Peritus
$
\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j 1 = } \sum\limits_{j = 1}^n {j = \left( {1 + n} \right)\frac{n}
{2}}
$

$\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^j {ij} = \sum\limits_{j = 1}^n {j\left( {1 + j} \right)\frac{j}
{2} = } } \frac{1}
{2}\left[ {\sum\limits_{j = 1}^n {j^2 + } \sum\limits_{j = 1}^n {j^3 } } \right]$

these are well known sums Power Sum -- from Wolfram MathWorld
How did you get to this

$\sum\limits_{j = 1}^n {j\left( {1 + j} \right)\frac{j}
{2} }$