TAOCP 1.2.3 Exercise 31
Apply Binet's formula from Exercise 30 with the substitutions a_j=u_j,\qquad b_j=1,\qquad x_j=1,\qquad y_j=v_j.
Section 1.2.3: Sums and Products
Exercise 31. [M20] Use Binet’s formula to express the sum $\sum_{1 \le j < k \le n} (u_j - u_k)(v_j - v_k)$ in terms of $\sum_{j=1}^n u_j v_j$, $\sum_{j=1}^n u_j$, and $\sum_{j=1}^n v_j$.
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Solve time: 1m06s
Solution
Apply Binet's formula from Exercise 30 with the substitutions
$$ a_j=u_j,\qquad b_j=1,\qquad x_j=1,\qquad y_j=v_j. $$
Then
$$ \left(\sum_{j=1}^n u_j\right)\left(\sum_{j=1}^n v_j\right) = \left(\sum_{j=1}^n u_jv_j\right)\left(\sum_{j=1}^n 1\right) + \sum_{1\le j<k\le n} (u_j-u_k)(v_j-v_k). $$
Since
$$ \sum_{j=1}^n 1=n, $$
this becomes
$$ \left(\sum_{j=1}^n u_j\right)\left(\sum_{j=1}^n v_j\right) = n\sum_{j=1}^n u_jv_j + \sum_{1\le j<k\le n} (u_j-u_k)(v_j-v_k). $$
Solving for the required sum yields
$$ \sum_{1\le j<k\le n} (u_j-u_k)(v_j-v_k) = n\sum_{j=1}^n u_jv_j - \left(\sum_{j=1}^n u_j\right) \left(\sum_{j=1}^n v_j\right). $$
Therefore
$$ \boxed{ \sum_{1\le j<k\le n} (u_j-u_k)(v_j-v_k) = n\sum_{j=1}^n u_jv_j - \left(\sum_{j=1}^n u_j\right) \left(\sum_{j=1}^n v_j\right) }. $$
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