Question

1 से 2001 तक की विषम पूर्णांकों का योग ज्ञात कीजिए​

Answer 1

Answer:

1,002,001

Step-by-step explanation:

$S_{n}$ = $\frac{n}{2}$ (2$a_{1}$ + (n - 1)d)

$a_{n}$ = $a_{1}$ + (n - 1)d

~~~~~~~~~~~~

1, 3, 5, ... , 2001

$a_{1}$ = 1 , d = 2 and $a_{n}$ = 2001

2001 = 1 + 2(n - 1)

2n - 1 = 2001

n = 1001

S = (1001/2)(2 + 2000) = 1001² = 1,002,001