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Find the sum of odd integers from 1 to 2001.

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Hint: Here, we will be using the formulas for ${n^{{\text{th}}}}$ term and the sum of first $n$ terms of an AP.

Since, the odd integers occurring from 1 to 2001 are $1,3,5,.....,1997,1999,2001$.
Here, ${a_1} = 1$, $d = 2$ and ${a_n} = 2001$(last term)
Clearly, the above series have a common difference of 2 and represents an arithmetic progression.
For an AP having the first term as ${a_1}$ and common difference as $d$, ${n^{{\text{th}}}}$ term of the AP is given by ${a_n} = {a_1} + \left( {n - 1} \right)d$
For the given series, ${n^{{\text{th}}}}$term (last term) of the AP is given by $2001 = 1 + \left( {n - 1} \right) \times 2 \Rightarrow 2001 = 1 + 2n - 2 \Rightarrow 2n = 2002 \Rightarrow n = 1001$
Therefore, the total number of terms in the given series is 1001.
Also, the formula for the sum of first $n$ terms of the AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$
So, the required sum of the given series is given by ${{\text{S}}_n} = \dfrac{{1001}}{2}\left[ {2\left( 1 \right) + \left( {1001 - 1} \right) \times 2} \right] = \dfrac{{1001}}{2}\left[ {2 + 2000} \right] = \dfrac{{1001 \times 2002}}{2} = 1002001$

Note: The difference between any two consecutive odd integers and between any two consecutive even integers is two because in between any two consecutive odd integers, one even integer occurs and similarly in between any two consecutive even integers, one odd integer occurs. The important thing in such types of problems is to identify the type of progression.
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