Derive a formula for evaluating the following expression in terms of n, given that n is a positive integer.
Σi = 1 to n2 ([√ i] + <√ i>)
Note: [x] is the greatest integer ≤ x, and <x> is the least integer ≥x.
Hmm. Wonder why nobody has tackled this yet? Must be Christmas break. Well, here's a start:
[√ i] = n if i = n*n, or n-1 if (n-1)(n-1) < i < n*n
<√ i> = n if (n-1)(n-1) < i <= n*n
so
[√ i] + <√ i> = 2n if i = n*n, or (2n-1) if (n-1)(n-1) < i < n*n
Going from (n-1)(n-1) to n*n involves adding
(n-1)(n-1) - n*n = 2n-1 terms,
1 whose value is 2n and (2n-2) whose value is (2n-1).
Simplifying, going from (n-1)(n-1) to n*n involves adding
2n + (2n-2)(2n-1) = 4n*n - 4n + 2
So, the sought after expression is the same as
Σi = 1 to n (4n*n - 4n + 2)
I'll stop here. These summations are fairly well known, but it's too much algebra for me right now (plus I am obviously not good at writing exponents on perplexus. The solution involves 4th powers, and I would embarrass myself writing n*n*n*n where I mean n to the 4th power).
a start (spoiler)
