I would like to add some info to these nice answers. If you deal with addition, then Paritosh answer is the simplest solution. Yet if your recursive function is more elaborate then addition, then you will probably need what Seth recommended. Documentation gives basic example how FoldList works with function f (note that the f itself is not recursive, - the recursion is provided by FoldList):
FoldList[f, x, {a, b, c, d}]
Out[] = {x, f[x, a], f[f[x, a], b], f[f[f[x, a], b], c], f[f[f[f[x, a], b], c], d]}
So the same idea applied to your case would look like
FoldList[Plus, a, {b, c, d}]
Out[] = {a, a + b, a + b + c, a + b + c + d}
which gives the same as
Accumulate[{a, b, c, d}]
{a, a + b, a + b + c, a + b + c + d}
And in a case of a function more complex than addition you need something like:
FoldList[#1^#2 &, x, {a, b, c, d}]
Out[] = {x, x^a, (x^a)^b, ((x^a)^b)^c, (((x^a)^b)^c)^d}
