Chapter 13: Q. 27 (page 1004)

Evaluate the sums in Exercises $23 - 28$ .
$\sum_{i = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3}$

Short Answer

Expert verified

The value of summation is $1260$

Step by step solution

Step 1. Given information

An expression is given as $\sum_{i = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3}$

Step 2. Evaluating summation

Separate out the terms as and simplify,

$\sum_{i = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3} = \sum_{i = 1}^{4} \sum_{j = 1}^{3} (\sum_{k = 1}^{2} i j^{2} k^{3}) \sum_{k = 1}^{2} i j^{2} k^{3} = i j^{2} 1^{3} + i j^{2} 2^{3} = i j^{2} (1^{3} + 2^{3}) = i j^{2} (1 + 8) = 9 i j^{2}$

Now solve every term similarly,

$\sum_{j = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3} = \sum_{j = 1}^{4} \sum_{j = 1}^{3} (\sum_{i = 1}^{2} j^{2} k^{3}) = \sum_{i = 1}^{4} (\sum_{j = 1}^{3} 9 i j^{2}) \sum_{j = 1}^{3} 9 i j^{2} = 9 i (\sum_{j = 1}^{3} j^{2}) = 9 i (1^{2} + 2^{2} + 3^{2}) = 9 i (1 + 4 + 9) = 9 i \times 14 = 126 i \sum_{i = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3} = \sum_{i = 1}^{4} (\sum_{j = 1}^{3} 9 i j^{2}) = \sum_{i = 1}^{4} (126 i) = 126 \times (\sum_{i = 1}^{4} (i)) = 126 \times (1 + 2 + 3 + 4) = 126 \times 10 = 1260$

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Evaluate the sums in Exercises $23 - 28$ .
$\sum_{i = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3}$

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.∑i=14∑j=13∑k=12ij2k3

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 $23 - 28$ .
$\sum_{i = 1}^{4} \sum_{j = 1}^{3} \sum_{k = 1}^{2} i j^{2} k^{3}$