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If lemmygrad is what op is commenting on, they left it out of the image and caption.

For the case that n = 0 (before the first run of the loop), x(0) = 1.

For the first actual case, n = 1. X(1) = x(0)*3*n = 1*3*1 = 3.

For the next case, n = 2. X(2) = x(1)*3*n = 3*3*2 = 18.

For the next case, n = 3. X(3) = x(2)*3*n = 18*3*3 = 162.

For the next and last case, n = 4. X(4) = 162*3*4 which I’m not computing. The computer value of x(4) is the value of the product loop.

If that doesn’t help, I could try helping again to rephrase, but I’m not sure what else to add.

I think it would be much better to write it in another language, but here’s another way to do the second one (this is on Visual Basic):

Dim n as long
Dim product as long
Product = 1

For n = 1 to 4
Product = product * 2 * n
Next n