2 Sum[A[i], {i, 1, n}] == Sum[2 A[i], {i, 1, n}]

With[{n = 5},
 Resolve[ForAll[Evaluate@Array[A, n],
   2 Sum[A[i], {i, 1, n}] == Sum[2 A[i], {i, 1, n}]]]]

2 Sum[A[i], {i, n}] == Sum[2 A[i], {i, n}] /. Sum[k_ x_, {y_, z_}] :> k Sum[x, {y, z}]

rule =  Sum[Times[Longest[u___], x___] , {y_, z_}] :>
         Times[u] Sum[Times[x], {y, z}] /; (FreeQ[{u}, y]);
 Sum[b[j] 2 A[i], {i, n}] /. rule