Let a is the first term and d is the common difference of the AP

Now given nth term = 2n + 1

First term a = 2*1 + 1 = 2+1 = 3          (Put n = 1)

Second term = 2*2 + 1 = 4+1 = 5        (Put n = 2)

d = 5 - 3 = 2

Sum of first n terms of AP = (n/2)*{2a + (n-1)d}

                                        = (n/2)*{2*3 + (n-1)2}

                                        = n*{3 + (n-1)}

                                        = n*{3 + n-1}

                                        = n*{n + 2}

                                        = n² + 2n  

So sum of first n terms of AP is n² + 2n