Obtaining numerical value from Recurrence Table

a[1] = 2;
a[n_] := a[n] = 4 Sum[a[i], {i, n - 1}]

Table[a[n], {n, 1, 10}]

{2, 8, 40, 200, 1000, 5000, 25000, 125000, 625000, 3125000}

You can introduce a memory variable b[n] and solve for both a[n] and b[n].

RecurrenceTable[{a[n + 1] == 4 b[n], b[n] == b[n - 1] + a[n], 
   a[1] == b[1] == 2}, {a, b}, {n, 1, 10}][[All, 1]]

(*   {2, 8, 40, 200, 1000, 5000, 25000, 125000, 625000, 3125000}   *)

Clear["Global`*"]

In your code, you have the wrong syntax for the Sum. However, even with the correct syntax

rt = RecurrenceTable[{a[n + 1] == 4 Sum[a[i], {i, 1, n}], a[1] == 2}, 
  a, {n, 1, 10}]

As stated in the error message, all instances of a[_] must have arguments of the form n + integer

Amplifying on the answer by Suba Thomas

a[1] = 2;
a[n_] := a[n] = 4 Sum[a[i], {i, n - 1}]

seq = Table[a[n], {n, 1, 10}]

(* {2, 8, 40, 200, 1000, 5000, 25000, 125000, 625000, 3125000} *)

You can use FindSequenceFunction to generalize from the sequence

y[n_] = FindSequenceFunction[seq, n]

a[200] == y[200]

(* True *)