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Here is a series: S = { $0,1,1,2,2,3,3 ...$}

To find out sum of $1st$ n terms, expression is: $$ S = \sum_{k=1}^{n}\left\lfloor\frac{k}{2}\right\rfloor $$ How do I simplify the above summation?

Also, I want to avoid fractions in the answer.

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up vote 2 down vote accepted

You do not need the first $0$ term when summing. Just consider the set $\{1,1,2,2,3,3,\dots\}$ then the sum of first $2n$ terms is $$ S_{2n} = 2 \sum_{i=1}^n i = 2 ( 1 + 2 + \dots + n) = n(n+1) $$ This can be seen by summing $1+2+\dots+n$ with $n+(n-1)+\dots+1$ and adding terms to terms. You get $n$ times the number $n+1$.

The sum of first $2n+1$ terms is $$S_{2n+1} = S_{2n} + n+1 = n(n+1)+n+1= (n+1)^2$$

Indeed when you compute $S_n$ you get $$ 1,2,4,6,9,12,16,\dots$$ which alternates between perfect squares terms and $n(n+1)$ terms.

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Well, that's exactly the thing we want to do, avoid fractions and remove the floor. So, what is the most natural thing to do when we want to remove the floor? Separate into cases! $$\sum_{k=1}^{n}\left \lfloor \frac{k}{2} \right \rfloor = \sum_{k=1}^{[n/2]} \left \lfloor \frac{2k}{2} \right \rfloor + \left \lfloor \frac{2k+1}{2} \right \rfloor = \sum_{k=1}^{[n/2]} (2k + 1/2) = t(t+1) + t/2$$ where $t = [n/2]$

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Deal separately with the cases $n$ even (say $n=2m$) and $n$ odd (say $n=2m+1$).

For the case $n=2m$, you want $2(1+2+3+\cdots)$, which is $m(m+1)$.

For the case $n=2m+1$, you get $m(m+1)+(m+1)$, which is $(m+1)^2$.

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