# How to evaluate the sum $\sum_{i=1}^{2n}\sum_{j=i}^{n+i}\sum_{k=1}^{j} n$?

I am trying to figure out the time complexity of the a given Algorithm, but using summations. Can anyone help get started on the inner summation, because I haven't come across such summation yet, and we just began solving these.

$$\sum_{i=1}^{2n}\sum_{j=i}^{n+i}\sum_{k=1}^{j} n$$

• Since the answer to the question does not depend on the algorithm, I think it is more appropriate to tag it to summation instad of algorithms and computational complexity since this isneither a question on algorithms nor on complexity theory. – Fawzy Hegab Jan 26 '17 at 19:45
• We were given an algorithm and going through the for loops, I obtained these summations, either way thanks. – Kono Dio Da Jan 26 '17 at 19:49

Consider the inner summation. Note that it is indexed by $k$, meaning that $k$ is variable in this summation while everything else is constant. In particular, $n$ is a constant. Therefore $$\sum_{k=1}^j n = n\sum_{k=1}^j1 = nj.$$

In light of the previous calculation, in the second summation we would like to compute

$$\sum_{j=i}^{n+i}\sum_{k=1}^j n = \sum_{j=i}^{n+1}nj = \sum_{j=1}^{n+1}nj-\sum_{j=1}^{i-1}nj.$$

Here the variable in the summation is $j$, so again we can pull $n$ in front to get

$$\sum_{j=1}^{n+1}nj-\sum_{j=1}^{i-1}nj = n \left(\sum_{j=1}^{n+i}j-\sum_{j=1}^{i-1}j\right) = n \left(\frac{(n+i)(n+i+1)}{2} - \frac{(i-1)i}{2}\right).$$

Note that a well known formula for summing the number from 1 to m (see https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_⋯)

Finally, the outer summation. We would now like to compute

\begin{align*} \sum_{i=1}^{2n}\sum_{j=i}^{n+i}\sum_{k=1}^j n &= \sum_{i=1}^{2n}n \left(\frac{(n+i)(n+i+1)}{2} - \frac{(i-1)i}{2}\right)\\ & = \sum_{i=1}^{2n}\frac{n^3+2n^2i+n^2+2ni}{2}. \end{align*}

In this summation, the variable is $i$ so we can factor out multiplicative factors of $n$. After breaking up the long fraction, this gives us

\begin{align*} &=\sum_{i=1}^{2n}\sum_{j=i}^{n+i}\sum_{k=1}^j n\\ &=\frac{n^3}{2}\left(\sum_{i=1}^{2n}1\right)+n^2\left(\sum_{i=1}^{2n}i\right)+\frac{n^2}{2}\left(\sum_{i=1}^{2n}1\right)+n\left(\sum_{i=1}^{2n}i\right)\\ & = \frac{n^3}{2}\left(2n\right)+n^2\left(\frac{2n(2n+1)}{2}\right)+\frac{n^2}{2}\left(2n\right)+n\left(\frac{2n(2n+1)}{2}\right), \end{align*} where I used the same formula to sum the number 1 to 2n, and I used this other well known formula $$\sum_{i=1}^n i^2 = \frac{i(i+1)(2i+1)}{6}.$$ for summing the numbers $1^2, 2^2,..., (2n)^2$ (see https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares ).

So I'm sure you'll want to reduce this, but the summation is above.

• Ah thank you very much! Those links will help me a lot. – Kono Dio Da Jan 26 '17 at 19:46
• In the question posted, the summation for $j$ runs from $\color{red}i$ to $n+i$, not $1$ to $n+i$. – Hypergeometricx Jan 31 '17 at 11:31
• Thanks. I'll make the change – JSP Jan 31 '17 at 19:30

If you just want time complexity, a $O(n^k)$ for some $k$ might be good enough.

The following uses the trivial bound (for $n > 1$) $\sum_{i = 1}^n i^k \lt \sum_{i = 1}^n n^k = n^{k+1}$.

$\begin{array}\\ \sum_{i=1}^{2n}\sum_{j=i}^{n+i}\sum_{k=1}^{j} n &=\sum_{i=1}^{2n}\sum_{j=i}^{n+i}jn\\ &=\sum_{i=1}^{2n}n\sum_{j=i}^{n+i}j\\ &<\sum_{i=1}^{2n}n(n+i)^2\\ &<\sum_{i=1}^{2n}n(n^2+2ni+i^2)\\ &=\sum_{i=1}^{2n}n^3+\sum_{i=1}^{2n}2n^2i+n\sum_{i=1}^{2n}i^2\\ &=2n^4+2n^2\sum_{i=1}^{2n}i+n\sum_{i=1}^{2n}i^2\\ &<2n^4+2n^2(2n)^2+n(2n)^3\\ &=(2+4+8)n^4\\ &=14n^4\\ \end{array}$

So the total is $O(n^4)$.

\begin{align} \sum_{i=1}^{2n}\sum_{j=i}^{n+i}\sum_{k=1}^j n &=n\sum_{i=1}^{2n}\sum_{j=i}^{n+i}j\\ &=n\sum_{i=1}^{2n}\sum_{r=0}^n (r+i) &&(r=j-i)\\ &=n\left[\sum_{i=1}^{2n}\sum_{r=0}^n r+\sum_{r=0}^n \sum_{i=1}^{2n}i\right]\\ &=n\left[2n\cdot \frac {n(n+1)}2+(n+1)\cdot \frac {(2n)(2n+1)}2\right]\\ &=\color{red}{n^2 (n+1)(3n+1)=\mathcal{O}(n^4)} \end{align}