This is just an extended comment as @MikeY has the answer.
For your particular example the function SymmetricReduction can help with seeing patterns:
k = 4;
bernoullies = Table[x[i] \[Distributed] BernoulliDistribution[p[i]], {i, k}];
Expectation[Log[1 + Sum[Log[1 + x[i]], {i, k}]], bernoullies];
SymmetricReduction[%, Table[p[i], {i, k}], Table[s[i], {i, k}]][[1]]
(* Log[1+Log[2]] s[1]+
(-2 Log[1+Log[2]]+Log[1+Log[4]]) s[2]+
(3 Log[1+Log[2]]- 3 Log[1+Log[4]]+Log[1+Log[8]]) s[3]+
(-4 Log[1+Log[2]]+ 6 Log[1+Log[4]]-4 Log[1+Log[8]]+Log[1+Log[16]]) s[4] *)
k = 5;
bernoullies = Table[x[i] \[Distributed] BernoulliDistribution[p[i]], {i, k}];
Expectation[Log[1 + Sum[Log[1 + x[i]], {i, k}]], bernoullies];
SymmetricReduction[%, Table[p[i], {i, k}], Table[s[i], {i, k}]][[1]]
(* Log[1+Log[2]] s[1]+
(-2 Log[1+Log[2]]+Log[1+Log[4]]) s[2]+
(3 Log[1+Log[2]]-3 Log[1+Log[4]]+Log[1+Log[8]]) s[3]+
(-4 Log[1+Log[2]]+6 Log[1+Log[4]]-4 Log[1+Log[8]]+Log[1+Log[16]]) s[4]+
(5 Log[1+Log[2]]-10 Log[1+Log[4]]+10 Log[1+Log[8]]-5 Log[1+Log[16]]+Log[1+Log[32]]) s[5] *)
k = 6;
bernoullies = Table[x[i] \[Distributed] BernoulliDistribution[p[i]], {i, k}];
Expectation[Log[1 + Sum[Log[1 + x[i]], {i, k}]], bernoullies];
SymmetricReduction[%, Table[p[i], {i, k}], Table[s[i], {i, k}]][[1]]
(* Log[1+Log[2]] s[1]+
(-2 Log[1+Log[2]]+Log[1+Log[4]]) s[2]+
(3 Log[1+Log[2]]- 3 Log[1+Log[4]]+Log[1+Log[8]]) s[3]+
(-4 Log[1+Log[2]]+ 6 Log[1+Log[4]]-4 Log[1+Log[8]]+Log[1+Log[16]]) s[4]+
(5 Log[1+Log[2]]-10 Log[1+Log[4]]+10 Log[1+Log[8]]- 5 Log[1+Log[16]]+Log[1+Log[32]]) s[5]+
(-6 Log[1+Log[2]]+15 Log[1+Log[4]]-20 Log[1+Log[8]]+15 Log[1+Log[16]]- 6 Log[1+Log[32]]+Log[1+Log[64]]) s[6] *)
From the observed pattern one can put together a general formula:
mean[k_] := Sum[SymmetricPolynomial[j, Table[p[i], {i, k}]] *
Sum[(-1)^(j - i) Binomial[j, i] Log[1 + Log[2^i]], {i, 1, j}], {j, 1, k}]
mean[4]
(* (-4 Log[1+Log[2]]+6 Log[1+Log[4]]-4 Log[1+Log[8]]+Log[1+Log[16]])
p[1] p[2] p[3] p[4]+ Log[1+Log[2]] (p[1]+p[2]+p[3]+p[4])+
(-2 Log[1+Log[2]]+Log[1+Log[4]]) (p[1] p[2]+p[1] p[3]+p[2] p[3]+
p[1] p[4]+p[2] p[4]+p[3] p[4])+(3 Log[1+Log[2]]-
3 Log[1+Log[4]]+Log[1+Log[8]]) (p[1] p[2] p[3]+p[1] p[2] p[4]+p[1] p[3] p[4]+
p[2] p[3] p[4]) *)
Use of the general formula is hundreds to thousands times faster than using Expectation.
ggin his answer to clarify things. (And you should avoid usingSubscriptexcept maybe for display purposes and use something likex[k]andp[k].) Also, I think that @MikeY 's answer is the answer. – JimB Apr 07 '20 at 21:42