reserve k, m, n, p, K, N for Nat;
reserve i for Integer;
reserve x, y, eps for Real;
reserve seq, seq1, seq2 for Real_Sequence;
reserve sq for FinSequence of REAL;

theorem Th6:
  n>0 implies bseq(k+1).n=(1/(k+1))*(bseq(k).n)*(aseq(k).n)
proof
  assume
A1: n>0;
  thus bseq(k+1).n = (n choose (k+1))*(n ^ (-(k+1))) by Def2
    .= ((n-k)/(k+1))*(n choose k)*(n ^ (-(k+1))) by Th5
    .= ((n-k)/(k+1))*(n choose k)*((n ^ (-k))/n) by A1,Th4
    .= (n-k)*(k+1)"*(n choose k)*((n ^ (-k))/n)
    .= (n-k)*(k+1)"*(n choose k)*((n ^ (-k))*n")
    .= (k+1)"*((n choose k)*(n ^ (-k)))*((n-k)*n")
    .= (1/(k+1))*((n choose k)*(n ^ (-k)))*((n-k)*n")
    .= (1/(k+1))*((n choose k)*(n ^ (-k)))*((n-k)/n)
    .= (1/(k+1))*(bseq(k).n)*((n-k)/n) by Def2
    .= (1/(k+1))*(bseq(k).n)*(aseq(k).n) by Def1;
end;
