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I should get sum of the following minimums.Is there any way to solve it?

$$\min\left\{2,\frac{n}2\right\} + \min\left\{3,\frac{n}2\right\} + \min\left\{4,\frac{n}2\right\} + \cdots + \min\left\{n+1, \frac{n}2\right\}=\sum_{i=1}^n \min(i+1,n/2)$$

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Just split the summation in two halves ($i+1 \leq n/2$ and $i+1 > n/2$) and evaluate both sums seperately. – TMM Sep 30 '11 at 21:11
I was wondering if the question is to prove the identity mentioned in the question. The word "solve" is abused! :-) – Srivatsan Sep 30 '11 at 21:16
up vote 2 down vote accepted

Suppose first that $n$ is even, say $n=2m$. Then $$\min\left\{i,\frac{n}2\right\}=\min\{i,m\}=\begin{cases}i,&\text{if }i\le m\\ m,&\text{if }i\ge m. \end{cases}$$

Thus, $$\begin{align*} \sum_{i=2}^{n+1}\min\left\{i,\frac{n}2\right\} &= \sum_{i=2}^m i + \sum_{i=m+1}^{n+1} m\\ &= \frac{m(m+1)}2-1 + (n+1-m)m\\ &= \frac12\left(\frac{n}2\right)\left(\frac{n}2+1\right)-1+\left(\frac{n}2\right)\left(\frac{n}2+1\right)\\ &= \frac{3n(n+2)}{8}-1\\ &=\frac{3n^2+6n-8}8. \end{align*}$$

If $n$ is odd, say $n=2m+1$, $$\min\left\{i,\frac{n}2\right\}=\min\left\{i,m+\frac12\right\}=\begin{cases}i,&\text{if }i\le m\\ m+\frac12,&\text{if }i> m. \end{cases}$$

Thus, $$\begin{align*} \sum_{i=2}^{n+1}\min\left\{i,\frac{n}2\right\} &= \sum_{i=2}^m i + \sum_{i=m+1}^{n+1} \left(m+\frac12\right)\\ &= \frac{m(m+1)}2-1 + (n+1-m)\left(m+\frac12\right)\\ &= \frac12\left(\frac{n-1}2\right)\left(\frac{n+1}2\right)-1+\left(\frac{n}2\right)\left(\frac{n+3}2\right)\\ &= \frac{3n^2+6n-1}{8}-1\\ &= \frac38 (n^2+2n-3). \end{align*}$$

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Since you've solved it completely: one could unify the even and odd cases as $\frac{3n^2}{8}+\frac{3n}{4}-1-\frac{1-(-1)^n}{16}$ for $n > 1$. The "odd formula" is zero when $n=1$, while the sum evaluates to $\frac12$. – J. M. Sep 30 '11 at 21:54

If $n$ is even, your sum splits as

$$\sum_{i=1}^{\frac{n}{2}-2} \min\left(i+1,\frac{n}{2}\right)+\frac{n}{2}+\sum_{i=\frac{n}{2}}^{n} \min\left(i+1,\frac{n}{2}\right)=\sum_{i=1}^{\frac{n}{2}-2} (i+1)+\frac{n}{2}+\frac{n}{2}\sum_{i=\frac{n}{2}}^{n} 1$$

If $n$ is odd, you can perform a similar split:

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

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