If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
an= a + (n–1) d = 2n + 1 (given)…………………(1)
Sn= n/2 (2a + (n–1) d)
By putting n=1 in (1)
a=3
similarly a2= 5
a3= 7
d= common difference = a2 – a1= 2
Sn = n/2 (6 + (n–1) 2)
= n/2(2n + 4)
= n (n + 2)