calculate-1-i-n-and-1-j-n-min-i-j-

Question Number 73028 by mathmax by abdo last updated on 05/Nov/19

c a l c u l a t e \sum_{1 ⩽ i ⩽ n a n d 1 ⩽ j ⩽ n} m i n (i, j)

Answered by mind is power last updated on 05/Nov/19

= \underset{i = 1}{\sum^{n}} i + 2 \underset{i = 2}{\sum^{n}} \underset{j = 1}{\sum^{i - 1}} (j)

= \underset{i = 1}{\sum^{n}} i + 2 \underset{i = 2}{\sum^{n}} . \frac{(i - 1) . i}{2} = \underset{i = 1}{\sum^{n}} i - \underset{i = 2}{\sum^{n}} i + \underset{i = 2}{\sum^{n}} i^{2} = 1 + \underset{i = 2}{\sum^{n}} i^{2} = \underset{i = 1}{\sum^{n}} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}

Answered by mr W last updated on 05/Nov/19

= 2 \times [n \times 1 + (n - 1) \times 2 + \dots + 1 \times n] - (1 + 2 + \dots + n)

= 2 \times \underset{k = 0}{\sum^{n - 1}} (n - k) (k + 1) - (1 + 2 + \dots + n)

= 2 \times \underset{k = 0}{\sum^{n - 1}} [(n - 1) k + n - k^{2}] - (1 + 2 + \dots + n)

= 2 \times [(n - 1) \frac{n (n - 1)}{2} + n^{2} - \frac{(n - 1) n (2 n - 1)}{6}] - \frac{n (n + 1)}{2}

= \frac{n (n + 1) (2 n + 1)}{6}

Commented by mathmax by abdo last updated on 05/Nov/19

t h a n k y o u s i r .