TADM2E 2.3

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f(n) = (((n^2)(n+1)^2)/8) + n(n+1)(2n+1)/12


This problem does appear to break down into a series of nested summations:

<math> \displaystyle\sum_{i=1}^{n}\text{ } \displaystyle\sum_{j=1}^{i}\text{ } \displaystyle\sum_{k=j}^{i+j}\text{ } \displaystyle\sum_{l=1}^{j+i-k}1 </math>

In the last summation, the formula is independent of the iterator, which translates into adding the value 1, <math>j+i-k</math> times:

<math> \displaystyle\sum_{i=1}^{n} \displaystyle\sum_{j=1}^{i} \displaystyle\sum_{k=j}^{i+j}(j+i-k) </math>

Now the third summation goes from <math>j</math> to <math>i+j</math> the formula on closer examination reveals that

<math>\displaystyle\sum_{k=j}^{i+j}(j+i-k)</math> is <math> \displaystyle\sum_{k=1}^{i}(k) </math> which is equal to <math>i*(i+1)/2</math>

So the summation boils down to

<math> \displaystyle\sum_{i=1}^{n} \displaystyle\sum_{j=1}^{i} (i*(i+1)/2) </math>

The formula in the second summation is independent of the iterator, which translates to adding <math>i*(i+1)/2</math>, <math>i</math> times.

<math> \displaystyle\sum_{i=1}^{n}(i^2 * (i+1)/2) </math>

which is

<math> \displaystyle\sum_{i=1}^{n}((i^3 + i^2)/2) </math>


<math>\frac{1}{2} \left(\Sigma r^3 + \Sigma r^2\right) = \frac{1}{2}\left(\frac{n^2\left(n+1\right)^2}{4} + \frac {n \left(n+1\right)\left(2n+1\right)}{6}\right) = \frac{n^2\left(n+1\right)^2}{8} + \frac {n \left(n+1\right)\left(2n+1\right)}{12}</math>


Time Complexity = O<math>({n}^{4})</math>