# How do you add all the odd numbers between 1-99 inclusive?

Sep 7, 2015

${\sum}_{n = 1}^{50} \left(2 n - 1\right) = 1 + 3 + 5 + 7 + 9 + \ldots \ldots = 97 + 99 = 2500$

#### Explanation:

Form a sequence $\left({x}_{n}\right) = 1 , 3 , 5 , 7 , 9 , \ldots \ldots , 95 , 97 , 99.$
This is the sequence of all the odd numbers between 1 and 99, endpoints included.

Clearly this is an arithmetic sequence with common difference d = 2 between terms.

The general term for this sequence may be given as :

${x}_{n} = a + \left(n - 1\right) d$ , where a = first term, n = number of terms.

Hence, ${x}_{n} = 1 + \left(n - 1\right) \left(2\right)$
$= 2 n - 1$

Writing ${x}_{n} = 99$ and solving for $n$ gives that 99 is the 50th term.

Now to add all the terms of this sequence, yields the arithmetic series
${\sum}_{n = 1}^{50} \left(2 n - 1\right) = 1 + 3 + 5 + 7 + 9 + \ldots \ldots = 97 + 99.$

From the formula for sum of an arithmetic series with constant difference d, we obtain the solution as
${\sum}_{n = 1}^{n} \left[a + \left(n - 1\right) d\right] = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$
$= \frac{50}{2} \left[\left(2\right) \left(1\right) + \left(50 - 1\right) \left(2\right)\right]$

$= 2500$

Or, using the other formula if the first and last terms are known is:
sum$= \frac{n}{2} \left(a + l\right) = \frac{50}{2} \left(1 + 99\right) = 2500$

Jan 16, 2017

$2500$

#### Explanation:

More simply, considering the sum of the series

$1 + 3 + 5 + 7 + \ldots + 93 + 95 + 97 + 99$

Note that the numbers may be paired off $\left(1 + 99\right)$, $\left(3 + 97\right)$, $\left(5 + 95\right)$, each pair adding to $100$. There are $25$ such pairs. So the sum equals $2500$ $\left(25 \times 100\right)$.