JAMB - Mathematics (2006 - No. 42)
The sum of the first n positive integers is
1/2 n(n-1)
n(n+1)
n(n-1)
1/2 n(n+1)
Giải thích
Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)