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

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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 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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