Short Answer

$How many summands are in \sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} ?$

$Total number of summands in \sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} are 3360 .$

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Step 1. Given Information

$\sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k}$

Step 2. Finding summands

$\sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} = \sum_{i = 2}^{15} \sum_{j = 3}^{17} (\sum_{k = 4}^{19} \frac{i}{j + k}) The summation \sum_{k = 4}^{19} \frac{i}{j + k} can be expanded by replacing k with 4 and rewriting each term when k ends at 19 . \sum_{k = 4}^{19} \frac{i}{j + k} = \frac{i}{j + 4} + \frac{i}{j + 5} + . . . . . + \frac{i}{j + 19} The number of terms in above summation are 19 - 4 + 1 = 16 . \sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} = \sum_{i = 2}^{15} \sum_{j = 3}^{17} (\sum_{k = 4}^{19} \frac{i}{j + k}) = \sum_{i = 2}^{15} \sum_{j = 3}^{17} (\frac{i}{j + 4} + \frac{i}{j + 5} + . . . . . + \frac{i}{j + 19}) = \sum_{i = 2}^{15} (\sum_{j = 3}^{17} (\frac{i}{j + 4} + \frac{i}{j + 5} + . . . . . + \frac{i}{j + 19})) Similarly when expanded \sum_{j = 3}^{17} (\frac{i}{j + 4} + \frac{i}{j + 5} + . . . . . + \frac{i}{j + 19}) by replacing j from 3 to 17, the number of \frac{i}{j + 4} + \frac{i}{j + 5} + . . . . . + \frac{i}{j + 19} terms obtained are 17 - 3 + 1 = 15 . Similarly when expanded the last summation with respect to i from 2 to 15, the number of terms of terms obtained are 15 - 2 + 1 = 14 . Hence total number of summands are 16 \times 15 \times 14 = 3360 Total number of summands in \sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} are 3360 .$

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$How many summands are in \sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} ?$

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Step 1. Given Information

Step 2. Finding summands

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$How many summands are in \sum_{i = 2}^{15} \sum_{j = 3}^{17} \sum_{k = 4}^{19} \frac{i}{j + k} ?$