TADM2E 2.3

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

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

$ \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 $

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

$ \displaystyle\sum_{i=1}^{n} \displaystyle\sum_{j=1}^{i} \displaystyle\sum_{k=j}^{i+j}(j+i-k) $

Now the third summation goes from $ j $ to $ i+j $ the formula on closer examination reveals that

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

Since $ \displaystyle\sum_{k=start}^{end}(end-k) = \displaystyle\sum_{k=1}^{end-start}(k) $

So the summation boils down to

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

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

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

which is

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

$ \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} $

Time Complexity = O$ ({n}^{4}) $

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