Compute the close form for $\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (3i-1)$ as a polynomial in n. The closed form solution should not have a summation in it.

I am not used to working with multiple summations so I'm not sure if I erred in my process or if this is the correct solution:

$\sum_{i=1}^n \sum_{j=1}^n (3\sum_{k=1}^ni-1\sum_{k=1}^n)$

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

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

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

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

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

(3n^4 + 1/2n^3)


1 Answer 1


\begin{align}\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (3i-1) &= \sum_{i=1}^n (3i-1)\left[\sum_{j=1}^n \sum_{k=1}^n 1\right]\\ &=n^2\sum_{i=1}^n (3i-1)\end{align}

Can you finish off the problem?


As you work on summation over the $k$ index, notice that the expression that is being worked on is independent of $k$. Index $k$ shouldn't influence index $i$.

Remark: Another mistake is one of the $\frac32$ become $3$.

  • $\begingroup$ if the expression that you are referring to was substituting with the expression with the index i, it would then be correct as that expansion deals with the same indexes, right? $\endgroup$ Commented Mar 1, 2018 at 5:20
  • $\begingroup$ I was referring to $(3i-1)$, as we work with $k$ index, $3i-1$ is being treated as a constant. I am not sure whether I am answering your latest question. $\endgroup$ Commented Mar 1, 2018 at 5:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .