Find an easy way to evaluate the following series: #3+5+7+cdots+ 17+19+21#?

Question

Nova Allen · Accepted Answer

#120#

Note that as you work inwards, groups of #24# form.

Add the first and last numbers, #3# and #21#, to get #24#.

Since the next two numbers inwards are just #2# less and greater than the numbers we added before, their sum will also be #24#. That is:

#5+19=(3+2)+(21-2)=(3+21)+(2-2)=24#

The pattern continues:

#{(3+21=24),(5+19=24),(7+17=24),(9+15=24),(11+13=24):}#

Thus, the sum is #5xx24=120#.

Genesis Amaya · Answer

undefined The sum of the series $3 + 5 + 7 + \dots + 17 + 19 + 21$ can be evaluated using the arithmetic series formula:

$$ S = \frac{n}{2} \cdot (a + l) $$

Where:
- $ S $ is the sum of the series.
- $ n $ is the number of terms in the series.
- $ a $ is the first term.
- $ l $ is the last term.

In this series:
- $ a = 3 $
- $ l = 21 $
- To find $ n $, we can use the formula for the $ n $th term of an arithmetic series: $ a_n = a + (n - 1) \cdot d $, where $ a_n $ is the $ n $th term and $ d $ is the common difference.

For the last term $ l = a + (n - 1) \cdot d $:
$$ 21 = 3 + (n - 1) \cdot 2 $$
$$ 18 = (n - 1) \cdot 2 $$
$$ 9 = n - 1 $$
$$ n = 10 $$

Now, we can use the formula for the sum:
$$ S = \frac{n}{2} \cdot (a + l) $$
$$ S = \frac{10}{2} \cdot (3 + 21) $$
$$ S = \frac{10}{2} \cdot 24 $$
$$ S = 5 \cdot 24 $$
$$ S = 120 $$

So, the sum of the series is $ 120 $.