Question

How do you calculate the sum of odd numbers between 1 and 2001 ?

Answer 1

Hint: In this question, we need to find the sum of odd numbers between 1 and 2001. Odd numbers are those which give fractional form when divided by 2. To find the sum, we make use of arithmetic progression related formulas. Firstly, we will find how many odd numbers came in between 1 and 2001. After finding it, we will calculate the sum of odd numbers between 1 and 2001.
The following formulas are used.
(1) ${a_n} = a + (n - 1)d$
(2) $S = \dfrac{n}{2}[a + {a_n}]$

Complete step-by-step solution:
In this question, we are asked to calculate the sum of odd numbers between 1 and 2001.
Note that odd numbers are those which give fractional form when divided by 2. Suppose we take the case of natural numbers, then the odd numbers among them will be given by $1,3.5,7,.....$
After every odd number, there comes an even number. These are placed alternatively in mathematics.
Firstly, we will list the odd numbers that come in between 1 and 2001.
They are given by, $3,5,7,.....,1999$
By using any arithmetic progression, we can easily find the sum of odd numbers.
We will use the formula of finding the nth term of an arithmetic progression which is given by,
${a_n} = a + (n - 1)d$ …… (1)
where $a = $ the first term of the sequence
             $n = $ the total numbers in the sequence
              $d = $ common difference
We find common difference using $d = {a_n} - {a_{n - 1}}$
In the given problem, ${a_1} = a = 3$, ${a_2} = 5$, ${a_3} = 7$ and ${a_n} = 1999$
Hence $d = {a_2} - {a_1} = {a_3} - {a_2}$
$ \Rightarrow d = 5 - 3 = 7 - 5$
$ \Rightarrow d = 2$
By using the equation (1) we find the total numbers in the sequence.
Hence by (1) we have,
$1999 = 3 + (n - 1)2$
Simplifying we get,
$ \Rightarrow 1999 = 3 + 2n - 2$
Rearranging the terms we get,
$ \Rightarrow 1999 = 2n + 1$
$ \Rightarrow 1999 - 1 = 2n$
$ \Rightarrow 1998 = 2n$
Taking 2 to the other side we get,
$ \Rightarrow n = \dfrac{{1998}}{2}$
$ \Rightarrow n = 999$
Now the sum of odd numbers is found using the formula,
$S = \dfrac{n}{2}[a + {a_n}]$ …… (2)
where $a = $ the first term of the sequence
             $n = $ the total numbers in the sequence
            ${a_n} = $ nth term in the sequence
Here $a = 3,$ $n = 999$ and ${a_n} = 1999$
Substituting these values in the equation (2), we get,
$ \Rightarrow S = \dfrac{{999}}{2}[3 + 1999]$
Simplifying we get,
$ \Rightarrow S = \dfrac{{999}}{2}[2002]$
$ \Rightarrow S = 999 \times 1001$
$ \Rightarrow S = 999999$
Hence, the sum of odd numbers between 1 and 2001 is $999999$.

Note: An arithmetic progression is a list of numbers in which each term is got by adding a fixed number to the preceding term except the first term.
The fixed number is called the common difference of the arithmetic progression and it is denoted by d. Note that the common difference d can be positive, negative or zero.
If ${a_1},{a_2},{a_3},{a_4},{a_5},....,{a_{n - 1}},{a_n}$ is an arithmetic progression, then we note that
$d = {a_2} - {a_1} = {a_3} - {a_2} = {a_4} - {a_3} = .... = {a_n} - {a_{n - 1}}.$
In order to determine whether a list of numbers is an arithmetic progression or not we need to just find the common difference d between the two numbers. If d is constant for all the terms then it is an arithmetic sequence. Otherwise it is not an arithmetic sequence.
We will use the formula of finding the nth term of an arithmetic progression which is given by,
${a_n} = a + (n - 1)d$
The sum of odd numbers is found using the formula,
$S = \dfrac{n}{2}[a + {a_n}]$