reconsider n = n as Element of NAT by ORDINAL1:def 13;
defpred S₁[ Nat, set ] means for l, m being Nat st m = $1 - 1 & l = n - m holds
$2 = ((n choose m) * (a |^ l)) * (b |^ m);
A4: for k being Nat st k in Seg (n + 1) holds
ex x being set st S₁[k,x]

proof

let k be Nat; :: thesis: ( k in Seg (n + 1) implies ex x being set st S₁[k,x] )
assume k in Seg (n + 1) ; :: thesis: ex x being set st S₁[k,x]
then A5: ( k >= 1 & k <= n + 1 ) by FINSEQ_1:3;
then reconsider m1 = k - 1 as Element of NAT by INT_1:18;
k - 1 <= (n + 1) - 1 by A5, XREAL_1:11;
then reconsider l1 = n - m1 as Element of NAT by INT_1:18;
take z = ((n choose m1) * (a |^ l1)) * (b |^ m1); :: thesis: S₁[k,z]
thus S₁[k,z] ; :: thesis: verum

end;

consider p being FinSequence such that
A6: dom p = Seg (n + 1) and
A7: for k being Nat st k in Seg (n + 1) holds
S₁[k,p . k] from FINSEQ_1:sch 1(A4);
rng p c= REAL

proof

let x be set ; :: according to TARSKI:def 3 :: thesis: ( not x in rng p or x in REAL )
assume x in rng p ; :: thesis: x in REAL
then consider y being set such that
A8: ( y in dom p & x = p . y ) by FUNCT_1:def 5;
reconsider y = y as Element of NAT by A8;
A9: ( y >= 1 & y <= n + 1 ) by A6, A8, FINSEQ_1:3;
then reconsider m1 = y - 1 as Element of NAT by INT_1:18;
y - 1 <= (n + 1) - 1 by A9, XREAL_1:11;
then reconsider l1 = n - m1 as Element of NAT by INT_1:18;
p . y = ((n choose m1) * (a |^ l1)) * (b |^ m1) by A6, A7, A8;
then reconsider x = x as Real by A8;
x in REAL ;
hence x in REAL ; :: thesis: verum

end;

then reconsider p = p as FinSequence of REAL by FINSEQ_1:def 4;
take p ; :: thesis: ( len p = n + 1 & ( for i, l, m being Nat st i in dom p & m = i - 1 & l = n - m holds
p . i = ((n choose m) * (a |^ l)) * (b |^ m) ) )
thus ( len p = n + 1 & ( for i, l, m being Nat st i in dom p & m = i - 1 & l = n - m holds
p . i = ((n choose m) * (a |^ l)) * (b |^ m) ) ) by A6, A7, FINSEQ_1:def 3; :: thesis: verum