The sum of all absolute values of the differe | vedclass

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(C) The sum is given by $S = \sum \limits_{1 \leq j < i \leq n} (i-j)$.We can rewrite this sum by counting how many times each difference $k = i-j$ oc

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${ }^{n+1} C_3$

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(C) The sum is given by $S = \sum \limits_{1 \leq j We can rewrite this sum by counting how many times each difference $k = i-j$ occurs.For a fixed difference $k$,where $1 \leq k \leq n-1$,the pairs $(j, i)$ such that $i-j = k$ are $(1, 1+k), (2, 2+k), \ldots, (n-k, n)$.There are exactly $(n-k)$ such pairs.Thus,the total sum is $S = \sum \limits_{k=1}^{n-1} k(n-k)$.$S = n \sum \limits_{k=1}^{n-1} k - \sum \limits_{k=1}^{n-1} k^2$.Using the formulas $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$ and $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$ with $m = n-1$:$S = n \frac{(n-1)n}{2} - \frac{(n-1)n(2n-2+1)}{6} = \frac{n^2(n-1)}{2} - \frac{n(n-1)(2n-1)}{6}$.$S = \frac{n(n-1)}{6} [3n - (2n-1)] = \frac{n(n-1)(n+1)}{6}$.Since ${ }^{n+1} C_3 = \frac{(n+1)n(n-1)}{3 	imes 2 	imes 1} = \frac{n(n-1)(n+1)}{6}$,the sum equals ${ }^{n+1} C_3$.

The sum of all absolute values of the differences of the numbers $1, 2, 3, \ldots, n$,taken two at a time,i.e.,$\sum \limits_{1 \leq j < i \leq n} |i-j|$ equals:

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