The sum of the products of the 2 n numbers ±1, ±2, ±3. …. ±n taking two at a time is (A) (n(n+1)

step 1 In this question, we have 1 minus 1 plus 2 minus 2 plus 3 minus 3 minus 3 plus and so on, 3 minu

The sum of the products of the 2 n numbers ±1, ±2, ±3. …. ±n...

The sum of the products of the $2 n$ numbers $\pm 1, \pm 2, \pm 3$. $\ldots . \pm n$ taking two at a time is
(A) $\frac{n(n+1)}{2}$
(B) $-\frac{n(n+1)}{2}$
(C) $\frac{n(n+1)(2 n+1)}{6}$
(D) $-\frac{n(n+1)(2 n+1)}{6}$

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Key Concepts

Sum of Squares Formula

The sum of the squares of the first n positive integers is a fundamental result typically expressed as n(n+1)(2n+1)/6. In contexts involving symmetric sets where both positive and negative counterparts are present, each square appears twice. Recognizing and correctly applying this formula is crucial when substituted into identities, such as the pairwise product sum formula, to achieve an exact result.

Pairwise Product Sum Formula

This concept involves recognizing that the sum of the products of all distinct pairs from a set can be efficiently calculated using the identity: (?a?)² = ?a?² + 2?a?a? (where the second sum runs over all distinct pairs). By rearranging the formula, one can express the pairwise products’ sum as ½[(?a?)² - ?a?²]. This approach connects the computation of pairwise products directly to the more straightforward sums of the numbers and their squares.

Symmetry in Number Sets

This key idea leverages the fact that when a set of numbers includes both a number and its negative, many symmetric cancellation properties occur. Particularly, the total sum of such a symmetric set is zero, which greatly simplifies calculations related to sums and pairwise products because certain terms vanish, leaving only contributions from the squared terms.

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