Step by step solution to "Is there any way to solve it?min{2,n2}+min{3,n2}+min{4,n2}+⋯+min{n+1,n2}=∑i=1nmin(i+1,n2)"

Alivia Rojas

Answered question

2022-04-14

Is there any way to solve it?

min {2, \frac{n}{2}} + min {3, \frac{n}{2}} + min {4, \frac{n}{2}} + \dots + min {n + 1, \frac{n}{2}} = \sum_{i = 1}^{n} min (i + 1, \frac{n}{2})

shanna87mn

Beginner2022-04-15Added 19 answers

Suppose first that n is even, say

n = 2 m

. Then

min {i, \frac{n}{2}} = min {i, m} = {\begin{array}{cc} i & if i \leq m \\ m & if i \geq m \end{array}

Thus,

\sum_{i = 2}^{n + 1} min {i, \frac{n}{2}} = \sum_{i = 2}^{m} i + \sum_{i = m + 1}^{n + 1} m

= \frac{m (m + 1)}{2} - 1 + (n + 1 - m) m

= \frac{1}{2} (\frac{n}{2}) (\frac{n}{2} + 1) - 1 + (\frac{n}{2}) (\frac{n}{2} + 1)

= \frac{3 n (n + 2)}{8} - 1

= \frac{3 n^{2} + 6 n - 8}{8}

If n is odd, say

n = 2 m + 1

min {i, \frac{n}{2}} = min {i, m + \frac{1}{2}} = {\begin{array}{cc} i & if i \leq m \\ m + \frac{1}{2} & if i \geq m \end{array}

Thus,

\sum_{i = 2}^{n + 1} min {i, \frac{n}{2}} = \sum_{i = 2}^{m} i + \sum_{i = m + 1}^{n + 1} (m + \frac{1}{2})

= \frac{m (m + 1)}{2} - 1 + (n + 1 - m) (m + \frac{1}{2})

= \frac{1}{2} (\frac{n - 1}{2}) (\frac{n + 1}{2}) - 1 + (\frac{n}{2}) (\frac{n + 3}{2})

= \frac{3 n^{2} + 6 n - 1}{8} - 1

= \frac{3}{8} (n^{2} + 2 n - 3)

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