n = 3;
F = Integrate[(D[Subscript[u, n] = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(
\*SuperscriptBox[\((1 - t)\), \(3\)]
\*SuperscriptBox[\(t\), \(3\)]
\*SuperscriptBox[\(t\), \(k - 1\)]
\*SubscriptBox[\(a\), \(k\)]\)\), {t, 2}])^2, {t, 0, 1}] +
Subscript[a, n + 1] (Integrate[(Subscript[u, n] = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\(
\*SuperscriptBox[\((1 - t)\), \(3\)]
\*SuperscriptBox[\(t\), \(3\)]
\*SuperscriptBox[\(t\), \(k - 1\)]
\*SubscriptBox[\(a\), \(k\)]\)\))^2, {t, 0, 1}] - 1);
Table[D[F, Subscript[a, i]] == 0, {i, n + 1}]
