The statement #S_n# claims that, if you sum the first #n# numbers of the form #3n-1#, the result will be #n*(3n+1)/2#

So, the statement #S_1# simply refers to the case #n=1#. This means that on the left hand side you don't really sum on anything, having one number alone. On the right hand side, you evaluate #n*(3n+1)/2#, given #n=1#. Let's compute both sides:

Since the generic number is expressed by #3n-1#, if #n=1# we have #3*1-1=2#. The left hand side is #2#.

The right hand side evaluates to #1*(3*1+1)/2 = 4/2 = 2#. The equation holds.

Let's deal with #S_2#, i.e. make the same considerations as above, but with #n=2#.

We have two sum the first two numbers: we already know that the first number is 2. The second number is #3*2-1 = 5#. So, the left hand side equals #2+5 = 7#.

The right hand side evaluates to #2*(3*2+1)/2 = 6+1 = 7#. The equation holds.

Similarly, when #n=3#, the third number is #3*3-1 = 8#. The left hand side is #2+5+8 = 15#

The right hand side evaluates to #3*(3*3+1)/2 = 3 * 10/2 = 3*5 = 15#. The equation holds.