[Solved]: Sum number of times statement is executed in triple nested loop

The Wolfram Alpha answer isn’t quite correct – you’re off by one in each sum – the loop will iterate over ${1,…,n-1}$, not ${1,…,n}$, though other than that your reasoning is sound. What you should be asking W/A for is: $$ sum^{n-1}_{i=0}sum^{i-1}_{j=0}sum^{j-1}_{k=0} 1 $$ If you break the sums down bit by bit it gets a little easier to swallow, just start at the innermost sum and work outwards. It’s probably fairly obvious that $sum^{j-1}_{k=0}1 = j$, which breaks the triple-sum down into $sum^{n-1}_{i=0}sum^{i-1}_{j=0}j$. We have that $sum^{i-1}_{j=0}j = frac{1}{2}(i^2-i)$, so what was a double sum is now the difference of two sums, $frac{1}{2}(sum^{n-1}_{i=0}i^2 – sum^{n-1}_{i=0}i)$. The second term has already been worked out, and we have that $sum^{n-1}_{i=0}i^2 = frac{n(n-1)(2n-1)}{6}$. From there, you should be able to work out your answer; begin{align*} frac{1}{2}left(sum^{n-1}_{i=0}i^2 – sum^{n-1}_{i=0}iright) &= frac{1}{2}left(frac{n(n-1)(2n-1)}{6} – frac{3n(n-1)}{6} right) &= frac{1}{2}left((2n-1)frac{n(n-1)}{6} – 3frac{n(n-1)}{6} right) &= frac{n(n-1)(2n-4)}{12} &= frac{n(n-1)(n-2)}{6} end{align*} The results about the sum of the first $n-1$ numbers and the sum of the first $n-1$ square numbers are fairly standard ones – if you want some proofs of them then this page gives some nice arguments, though note that their sums end at $n$ and not $n-1$.