QuestionState whether the given sums are equal or unequal. (a) $\sum_{i=1}^{21} i$ and $\sum_{k=1}^{21} k$ are ? (b) $\sum_{i=1}^{21} i$ and $\sum_{i=5}^{25}(i-4)$ are $\square$ ? ) (c) $\sum_{i=4}^{21} i(i-4)$ and $\sum_{j=0}^{17}(j+4) j$ are $\square$ ? ( (d) $\sum_{i=4}^{21} i(i-4)$ and $\sum_{k=4}^{21}\left(k^{2}-4 k\right)$ are $\square$

Studdy Solution

STEP 1

1. The notation $\sum$ denotes a summation, which is the addition of a sequence of numbers.
2. The index of summation (e.g., $i$ , $k$ , $j$ ) is a dummy variable and does not affect the value of the sum.
3. We will evaluate each pair of sums to determine if they are equal or unequal.

STEP 2

1. Compare the sums in part (a).
2. Compare the sums in part (b).
3. Compare the sums in part (c).
4. Compare the sums in part (d).

STEP 3

Compare the sums $\sum_{i=1}^{21} i$ and $\sum_{k=1}^{21} k$ :
Since both sums have the same limits and the same expression being summed, they are equal. The index of summation is a dummy variable and does not affect the sum.

STEP 4

Compare the sums $\sum_{i=1}^{21} i$ and $\sum_{i=5}^{25}(i-4)$ :
Rewrite $\sum_{i=5}^{25}(i-4)$ by substituting $j = i - 4$ , which gives $i = j + 4$ . The limits change as follows: when $i = 5$ , $j = 1$ ; when $i = 25$ , $j = 21$ .
Thus, $\sum_{i=5}^{25}(i-4) = \sum_{j=1}^{21} j$ , which is equal to $\sum_{i=1}^{21} i$ .

STEP 5

Compare the sums $\sum_{i=4}^{21} i(i-4)$ and $\sum_{j=0}^{17}(j+4) j$ :
Rewrite $\sum_{j=0}^{17}(j+4) j$ by substituting $i = j + 4$ , which gives $j = i - 4$ . The limits change as follows: when $j = 0$ , $i = 4$ ; when $j = 17$ , $i = 21$ .
Thus, $\sum_{j=0}^{17}(j+4) j = \sum_{i=4}^{21} i(i-4)$ .

STEP 6

Compare the sums $\sum_{i=4}^{21} i(i-4)$ and $\sum_{k=4}^{21}\left(k^{2}-4 k\right)$ :
Notice that $\sum_{k=4}^{21}\left(k^{2}-4 k\right)$ is exactly the same as $\sum_{i=4}^{21} i(i-4)$ since $i(i-4) = i^2 - 4i$ .
The answers are: (a) Equal (b) Equal (c) Equal (d) Equal

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