The sum of all possible products taken two at a time out of the numbers \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5\) is
-55
Understanding the Problem: Sum of Products
This question asks us to find the sum of all possible products formed by taking pairs of numbers from the set \(\{\pm 1, \pm 2, \pm 3, \pm 4, \pm 5\}\). This involves understanding how to calculate pairwise products and sum them up systematically.
Identifying the Set of Numbers
First, let's list the complete set of numbers provided:
\[ S = \{1, -1, 2, -2, 3, -3, 4, -4, 5, -5\} \]
This set contains 10 numbers in total. Let's denote these numbers as \(x_1, x_2, \dots, x_{10}\).
Strategy: Using Algebraic Identity
A direct calculation of all possible products (there would be \(\binom{10}{2} = \frac{10 \times 9}{2} = 45\) pairs) can be tedious. A more efficient method is to use the algebraic identity related to the square of a sum:
\[ \left( \sum_{i=1}^{n} x_i \right)^2 = \sum_{i=1}^{n} x_i^2 + 2 \sum_{1 \le i < j \le n} x_i x_j \]
We are looking for the value of \(\sum_{1 \le i < j \le 10} x_i x_j\). We can rearrange the formula as:
\[ \sum_{1 \le i < j \le 10} x_i x_j = \frac{1}{2} \left[ \left( \sum_{i=1}^{10} x_i \right)^2 - \sum_{i=1}^{10} x_i^2 \right] \]
To use this formula, we need to compute two sums: the sum of the numbers and the sum of the squares of the numbers.
Calculating the Sum of the Numbers
Let's calculate the sum of all the numbers in the set \(S\):
\[ \sum_{i=1}^{10} x_i = 1 + (-1) + 2 + (-2) + 3 + (-3) + 4 + (-4) + 5 + (-5) \]
Grouping the positive and negative terms:
\[ \sum_{i=1}^{10} x_i = (1 - 1) + (2 - 2) + (3 - 3) + (4 - 4) + (5 - 5) \]
\[ \sum_{i=1}^{10} x_i = 0 + 0 + 0 + 0 + 0 = 0 \]
The sum of all the numbers in the set is 0.
Calculating the Sum of the Squares of the Numbers
Next, we calculate the sum of the squares of each number in the set \(S\):
\[ \sum_{i=1}^{10} x_i^2 = (1)^2 + (-1)^2 + (2)^2 + (-2)^2 + (3)^2 + (-3)^2 + (4)^2 + (-4)^2 + (5)^2 + (-5)^2 \]
Remember that the square of a negative number is positive, so \((-a)^2 = a^2\).
\[ \sum_{i=1}^{10} x_i^2 = 1^2 + 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 4^2 + 4^2 + 5^2 + 5^2 \]
\[ \sum_{i=1}^{10} x_i^2 = 2 \times (1^2) + 2 \times (2^2) + 2 \times (3^2) + 2 \times (4^2) + 2 \times (5^2) \]
Factor out the 2:
\[ \sum_{i=1}^{10} x_i^2 = 2 \times (1^2 + 2^2 + 3^2 + 4^2 + 5^2) \]
Now, calculate the sum of the squares of the first 5 positive integers:
\[ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \]
Substitute this back:
\[ \sum_{i=1}^{10} x_i^2 = 2 \times 55 = 110 \]
The sum of the squares of all the numbers is 110.
Final Calculation of the Sum of Products
Now, substitute the calculated sums (\(\sum x_i = 0\) and \(\sum x_i^2 = 110\)) into the rearranged formula:
\[ \sum_{1 \le i < j \le 10} x_i x_j = \frac{1}{2} \left[ (0)^2 - 110 \right] \]
\[ \sum_{1 \le i < j \le 10} x_i x_j = \frac{1}{2} [0 - 110] \]
\[ \sum_{1 \le i < j \le 10} x_i x_j = \frac{1}{2} [-110] \]
\[ \sum_{1 \le i < j \le 10} x_i x_j = -55 \]
Conclusion
The calculation shows that the sum of all possible products taken two at a time from the numbers \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5\) is -55.
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