Question

please derive a closed formula of f(n) and please dont answer this question if you are not sure. because this is the second time I’m posting this question

Censiden MATRIX-eHAIN m p.kngth a ore 11 n 0 1 ton-t 1 and

Answer 1

The innermost loop with variable k, for a particular value of $l$ , evaluates from $i$ to  $j-1$ where $i$ range from 1 to  $n-l+1$ and $j=i+l-1$ .

Thus for a particular value of $i$ say $i = m, j=m+l-1$ , innermost loop will run for $j-1-i+1 = j-i = i+l-1 - i = l-1$ iterations.

And since $i$ range from 1 to  $n-l+1$ and for each value of $i$ , the value of $j$ is such that, innermost loop with variable k run for $l-1$ iterations.

Since there are $n-l+1$ possible values of $i$ in the range   1 to  $n-l+1$ and hence number of times innermost loop will execute for a particular value of $l$ will be $(n-l+1)(l-1)$ .

Since the value of $l$ range from $2$ to $n$,

Hence number of times, innermost loop with variable k will execute will be

$\sum_{l=2}^{n}(n-l+1)(l-1) = \sum_{x=1}^{n-1}(n-x)x \text{. where }x=l-1$

$\sum_{x=1}^{n-1}nx- \sum_{x=1}^{n-1}x^2 = n(n(n-1)/2)-n(n-1)(2n-1)/6 \\=n(n-1)(n+1)/6 = n(n^2-1)/6$

Please comment for any clarification.