Question

Find the sum of odd integers from 1 to 2001.

Answer 1

Hint: Here, we have to find the sum of odd integers from 1 to 2001. We will write the odd integers from 1 to 2001 in the form of an A.P. We will use the formula for \[{n^{{\rm{th}}}}\] term of an A.P. to find the number of terms in the A.P. Then, We will use the formula for the sum of \[n\] terms of an A.P. to find the sum of the odd integers from 1 to 2001.

Formula Used: The \[{n^{{\rm{th}}}}\] term of an A.P. is given by the formula \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term of the A.P. and \[d\] is the common difference.
The sum of \[n\] terms of an A.P. is given by the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], where \[a\] is the first term of the A.P. and \[d\] is the common difference.

Complete step-by-step answer:
We know that odd integers are the numbers not divisible by 2. For example: 1, 3, 5 are odd integers.
First, we will write the odd integers from 1 to 2001 in the form of a series.
Therefore, we get 1, 3, 5, 7, …, 1999, 2001.
We can observe that the series 1, 3, 5, 7, …, 1999, 2001 is an arithmetic progression with first term 1 and common difference \[3 - 1 = 2\].
Now, we will find the number of terms of A.P.
We can observe that the first term of the A.P. is 1, and the last term of the A.P. is 2001.
Also, we know that the \[{n^{{\rm{th}}}}\] term of an A.P. is given by the formula \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term of the A.P. and \[d\] is the common difference.
We will substitute the last term 2001 as the \[{n^{{\rm{th}}}}\] term, and solve the equation to find the number of terms in the A.P.
Substituting \[{a_n} = 2001\], \[a = 1\], and \[d = 2\] in the formula for \[{n^{{\rm{th}}}}\] term of an A.P., we get
\[ \Rightarrow 2001 = 1 + \left( {n - 1} \right)2\]
Subtracting 1 from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow 2001 - 1 = 1 + \left( {n - 1} \right)2 - 1\\ \Rightarrow 2000 = \left( {n - 1} \right)2\end{array}\]
Dividing both sides by 2, we get
\[\begin{array}{l} \Rightarrow \dfrac{{2000}}{2} = \dfrac{{\left( {n - 1} \right)2}}{2}\\ \Rightarrow 1000 = n - 1\end{array}\]
Adding 1 to both sides, we get
\[\begin{array}{l} \Rightarrow 1000 + 1 = n - 1 + 1\\ \Rightarrow 1001 = n\end{array}\]
Therefore, the number of terms in the A.P. 1, 3, 5, 7, …, 1999, 2001 is 1001.
Finally, we need to find the sum of all odd integers from 1 to 2001, that is \[1 + 3 + 5 + \ldots \ldots + 2001\].
We can observe that the sum \[1 + 3 + 5 + \ldots \ldots + 2001\] is the sum of all the terms of the A.P. 1, 3, 5, 7, …, 1999, 2001.
The sum of \[n\] terms of an A.P. is given by the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], where \[a\] is the first term of the A.P. and \[d\] is the common difference.
Substituting \[n = 1001\], \[a = 1\], and \[d = 2\] in the formula for sum of \[n\] terms of an A.P., we get
\[ \Rightarrow {S_{1001}} = \dfrac{{1001}}{2}\left[ {2\left( 1 \right) + \left( {1001 - 1} \right)2} \right]\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {S_{1001}} = \dfrac{{1001}}{2}\left[ {2 + \left( {1000} \right)2} \right]\\ \Rightarrow {S_{1001}} = \dfrac{{1001}}{2}\left( {2 + 2000} \right)\\ \Rightarrow {S_{1001}} = \dfrac{{1001}}{2}\left( {2002} \right)\\ \Rightarrow {S_{1001}} = 1001 \times 1001\end{array}\]
Multiplying 1001 by 1001, we get
\[ \Rightarrow {S_{1001}} = 1002001\]
\[\therefore\] The sum of all the odd integers from 1 to 2001 is 1,002,001.

Note: We can also find the sum using the last term of the A.P.
The sum of all terms of an A.P. is given by the formula \[{S_n} = \dfrac{n}{2}\left( {a + l} \right)\], where \[a\] is the first term of the A.P., \[l\] is the last term of the A.P. and \[n\] is the number of terms of the A.P.
Substituting \[n = 1001\], \[a = 1\], and \[l = 2001\] in the formula for sum of \[n\] terms of an A.P., we get
\[ \Rightarrow {S_{1001}} = \dfrac{{1001}}{2}\left( {1 + 2001} \right)\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {S_{1001}} = \dfrac{{1001}}{2}\left( {2002} \right)\\ \Rightarrow {S_{1001}} = 1001 \times 1001\end{array}\]
Multiplying 1001 by 1001, we get
\[ \Rightarrow {S_{1001}} = 1002001\]
\[\therefore\] The sum of all the odd integers from 1 to 2001 is 10,02,001.