# Find the sum of odd integers from $1$ to $2001$

$\begin{array}{1 1}(1000)^2 \\ (1001)^2 \\ (2001)^2 \\ 1001000 \end{array}$

Toolbox:
• $n^{th}$ term of an $A.P.$ is $t_n=a+(n-1)d$ where $a$ is first term and $d$ is common difference
• Sum of $n$ terms of an $A.P.$ is $S_n=\large\frac{n}{2}$$(a+l) where l is the last term Sum of odd integers from 1\:\:to\:\:2001 is 1+3+5+.................2001 The difference between any two successive terms is same =2 \therefore Clearly this series is an A.P. Here n^{th} term t_n=2001, a=1 and common difference d=2 We know that n^{th} term of an A.P. is t_n=a+(n-1)d \Rightarrow\:2001=1+(n-1)2 \Rightarrow\:n-1=1000 or n=1001 \therefore The sum of 1001 terms of the A.P. is S_{1001}=\large\frac{1001}{2}$$(1+2001)$
$i.e.,$ The sum of odd integers from $1\:\:to\:\:2001=1001\times1001=(1001)^2$