Summation over list elements

Question

As we know, one of the simplest way to do a summation (with special intervals) over list elements is using

Sum[list[[i]], {i, 3, 8, 2 }] (*with steps of 2*)

for a list such as list={a, b, c, d, e, f, g, h, i, j, k, l} which returns c + e + g.

How can we use Replacelist or FoldList to this aim?

I am learning important functions (FoldList and ReplaceList) which are used instead of (Accumulation) in sum over list, in which I faced to some problems and could not find correct things. Instead of accumulation, for a simple summation what does it work!?

Answer 1

I'd allow myself some guesswork here, that you're going after alternatives as a form of excercise in Mathematica programming. However, you've clearly stated, that the list is

list={a, b, c, d, e, f, g, h, i, j, k, l}

and the expected output is

c + e + g

Note, that in the question you've linked, ReplaceList and FoldList are utilized to generate lists of results, analogous to Accumulate. But even then, the functions used in the replacement rules or passed as arguments to FoldList are the conventional Plus. In this question, only a single expression is expected.

Off the top of my head I can only come up with this improvement to your code:

Total@list[[3;;8;;2]]
(* c + e + g *)