1. ## Double Summation question

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

i) $\displaystyle a_{ij}$ = 1
ii) $\displaystyle a_{ij}$ = 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:

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

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

It's this:

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

for 1 and ij.

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

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

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

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

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

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

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