If n^th term of an A.P. is 2n + 1, n sum of first n terms of A.P. is

$$\eqalign{ & {a_n} = 2n + 1 \cr & {a_1} = 2 \times 1 + \,1 = 2 + 1 = 3 \cr & {a_2} = 2 \times 2 + 1 = 4 + 1 = 5 \cr & \therefore d = {a_2} - {a_1} = 5 - 3 = 2 \cr & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2 \times 3 + \left( {n - 1} \right) \times 2} \right] \cr & = \frac{n}{2}\left[ {6 + 2n - 2} \right] \cr & = \frac{n}{2}\left[ {2n + 4} \right] \cr & = n\left[ {n + 2} \right] \cr} $$