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

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

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  • $n^{th}$ 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}$$(a+l)$ where $l$ is the last term
Sum of odd integers from $1\:\:to\:\:2001$ is
The difference between any two successive terms is same $=2$
$\therefore$ Clearly this series is an A.P.
Here $n^{th}$ term $t_n=2001$, $a=1$ and common difference $d=2$
We know that $n^{th}$ term of an A.P. is $t_n=a+(n-1)d$
$\Rightarrow\:n-1=1000$ or $ n=1001$
$\therefore$ The sum of 1001 terms of the A.P. is $S_{1001}=\large\frac{1001}{2}$$(1+2001)$
$i.e.,$ The sum of odd integers from $1\:\:to\:\:2001=1001\times1001=(1001)^2$
answered Feb 18, 2014 by rvidyagovindarajan_1

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