Series

First of all, we will need to use the following three series formulas to solve this problem. They are:

$\displaystyle \sum_{i=1}^n (i)=\frac{n(n+1)}{2} \\ \displaystyle \sum_{i=1}^n (i^2)=\frac{1}{6}n(n+1)(2n+1) \\ \displaystyle \sum_{i=1}^n (i^3)=(\frac{n(n+1)}{2})^2$ .

Also, we will use some basic summation manipulations which you find here in the "Identities" section.

Now, the given problem can be represented as follows:

$\displaystyle \large \sum_{i=1}^{22} \bigg( \sum_{j=1}^i (j^2) \bigg)$

$= \displaystyle \large \sum_{i=1}^{22} \bigg( \frac{1}{6}i(i+1)(2i+1) \bigg)$

$= \displaystyle \large \frac{1}{6} \centerdot \sum_{i=1}^{22} \bigg( 2i^3+3i^2+i \bigg)$

$= \displaystyle \large \frac{1}{6} \centerdot \bigg( 2\sum_{i=1}^{22} (i^3) + 3\sum_{i=1}^{22} (i^2) + \sum_{i=1}^{22} (i) \bigg)$

$= \displaystyle \frac{1}{6} \bigg( 2(\frac{22(22+1)}{2})^2 + 3(\frac{1}{6}\centerdot 22(22+1)(2\times 22 +1)) + \frac{22(22+1)}{2} \bigg)$

$= \displaystyle \frac{1}{6}(11\times 23)\bigg( 2(11\times 23) + 45 +1 \bigg)$

$= \displaystyle \frac{1}{6}(11\times 23)(22\times 23 + 46)$

$=\displaystyle \frac{46+506}{6} \times 11 \times 23$

$=92\times 11\times 23 = \boxed{23276}$

I solved this in a very simple fashion. First, we can note that there are 22 terms, and second, we can note that each term appears (23-n) times. With this in mind, we can write the following formula:

$\displaystyle \sum _{ n=1 }^{ 22 }{ (23-n)({ n }^{ 2 }) }$

$\displaystyle \sum _{ n=1 }^{ 22 }{ (23{ n }^{ 2 }-{ n }^{ 3 }) }$

$23\displaystyle \sum _{ n=1 }^{ 22 }{ { n}^{ 2 }-\sum _{ n=1 }^{ 22 }{ { n }^{ 3 } } }$

Now these become manageable. We know these two formulas:

$\displaystyle\sum _{ i=1 }^{ n }{ { i }^{ 2 } } =\frac { n(n+1)(2n+1) }{ 6 }$

$\displaystyle\sum _{ i=1 }^{ n }{ { i }^{ 3 } } ={ (\frac { n(n+1) }{ 2 } })^{ 2 }$

With these two formulas, all that is left is for us to evaluate:

$23\displaystyle \sum _{ n=1 }^{ 22 }{ { n}^{ 2 }-\sum _{ n=1 }^{ 22 }{ { n }^{ 3 } } }$ = $23(\frac { 22(22+1)(2(22)+1) }{ 6 }) -{ (\frac { 22(22+1) }{ 2 } })^{ 2 }$

After simplifying each expression we get:

$23(3795)-64009$ = $\boxed {23276}$