the proof

(%i1) s(n) := (2*a + (n-1)*d)*(n/2)$

(%i2) S1: s(n)$

(%i3) S2: s(2*n)$

(%i4) S3: s(3*n)$

(%i5) expand(S1);
                                  2
                               d n    d n
(%o5)                          ---- - --- + a n
                                2      2
(%i6) expand(S2);
                                  2
(%o6)                        2 d n  - d n + 2 a n
(%i7) expand(S3);
                                 2
                            9 d n    3 d n
(%o7)                       ------ - ----- + 3 a n
                              2        2
(%i8) expand(S2-S1);
                                   2
                              3 d n    d n
(%o8)                         ------ - --- + a n
                                2       2
(%i9) expand(3*(S2-S1));
                                 2
                            9 d n    3 d n
(%o9)                       ------ - ----- + 3 a n
                              2        2
(%i10) is(equal(s(3*n),3*(s(2*n)-s(n))));
(%o10)                               true


First, we find and expand the value of S1, S2 and S3 in the lines marked %o5, %o6 and %o7.
Then, in line %o8, we calculate the value of S2 - S1. Then we multiply it with 3 in line %o9.
Since, the value of %o7 and %o9 is the same, therefore S3=3(S2-S1)

%o10 is a one line shortcut in Maxima that does the whole thing together.