Chapter 13: Q. 26 (page 1004)

Evaluate the sums in Exercises $23 - 28$ .
$\sum_{j = 1}^{m} \sum_{k = 1}^{n} (j - k)$

Short Answer

Expert verified

The value of the summation is $\frac{1}{2} m n (m - n)$

Step by step solution

Step 1. Given information

An expression is given as $\sum_{j = 1}^{m} \sum_{k = 1}^{n} (j - k)$

Step 2. Evaluating summation

First separate out the term then simplify,

$\sum_{j = 1}^{m} \sum_{k = 1}^{m} (j - k) = \sum_{j = 1}^{m} (\sum_{k = 1}^{n} (j - k)) \sum_{k = 1}^{n} (j - k) = (j - 1) + (j - 2) + (j - 3) + \dots + (j - n) = n \times j - (1 + 2 + 3 + \dots + n) = n \times j - \frac{n (n + 1)}{2}$

Now simplify the other terms in the same manner,

$\sum_{j = 1}^{m} \sum_{k = 1}^{n} (j - k) = \sum_{j = 1}^{m} (\sum_{k = 1}^{n} (j - k)) = \sum_{j = 1}^{m} (n \times j - \frac{n (n + 1)}{2}) = (n \times 1 - \frac{n (n + 1)}{2}) + (n \times 2 - \frac{n (n + 1)}{2}) + (n \times 3 - \frac{n (n + 1)}{2}) + . . . . . . . + (n \times m - \frac{n (n + 1)}{2}) = n (1 + 2 + 3 + \dots + m) - m \times \frac{n (n + 1)}{2} = n \times \frac{m (m + 1)}{2} - m \times \frac{n (n + 1)}{2} = m n (\frac{m + 1}{2} - \frac{n + 1}{2}) = m n (\frac{m - n}{2}) = \frac{1}{2} m n (m - n)$

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Evaluate the sums in Exercises $23 - 28$ .
$\sum_{j = 1}^{m} \sum_{k = 1}^{n} (j - k)$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Evaluating summation

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Evaluate the sums in Exercises 23–28.∑j=1m∑k=1n(j-k)

Short Answer

Step by step solution

Step 1. Given information

Step 2. Evaluating summation

One App. One Place for Learning.

Most popular questions from this chapter

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Evaluate the sums in Exercises $23 - 28$ .
$\sum_{j = 1}^{m} \sum_{k = 1}^{n} (j - k)$