Question

find the sum of odd integers from 1 to 2001.

Answer 1

here is your answer.........

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Odd integers

= 1, 3 , 5 , 7 , 9 ....... , 2001

Above sequence is showing that the numbers are in Arithmetic Progression

$\textbf{\underline{First\;term=1}}$

$\textbf{\underline{Common\; Difference}}$

= 3 - 1

= 2

$\textbf{\underline{Using\;Formula}}$

$\tt{\rightarrow a_{n}=a+(n-1)d=2001}$

$\tt{\rightarrow a_{n}=1+(n-1)2=2001}$

2n - 2 = 2000

2n = 2000 + 2

2n = 2002

$\tt{\rightarrow n=\dfrac{2002}{2}}$

n = 1001

$\tt{\rightarrow S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

$\tt{\rightarrow S_{1001}=\dfrac{1001}{2}[2(1)+(1001-1)2]}$

$\tt{\rightarrow S_{1001}=\dfrac{1001}{2}[2+1000\times 2]}$

$\tt{\rightarrow S_{1001}=\dfrac{1001}{2}\times 2002}$

= 1001 × 1001

= 1002001