We have to find 1 + 3 + 5 + ......... + 2001 This is an A.P. with first term a = 1, common difference d = 3 – 1 = 2 and last term l = 2001 ∴ l = a + (n – 1)d ⇒ 2001 = 1 + (n – 1)2 ⇒ 2001 = 1 + 2n – 2 ⇒ 2n = 2001 + 1 ⇒ n = 1001 ⇒ Sum = $\frac{n}{2}$ (a + l) = $\frac{1001}{2}$ (1 + 2001) = (1001) (1001) = 1002001