reserve a,b,i,j,k,l,m,n for Nat;

theorem SumS:
  for a,b be Real, n be Nat holds
    a|^(n+1) - b|^(n+1) = (a-b)*Sum((a,b)Subnomial n)
proof
  let a,b be Real, n be Nat;
  A1: Sum ((a,b) Subnomial (n+1)) = Sum ((a*((a,b)Subnomial n))^<*b|^(n+1)*>)
    by RS
  .= Sum(a*((a,b)Subnomial n)) + b|^(n+1) by RVSUM_2:31
  .= a*Sum((a,b) Subnomial n) + b|^(n+1) by RVSUM_2:38;
  Sum ((a,b) Subnomial (n+1)) = Sum (<*a|^(n+1)*>^(b*((a,b)Subnomial n))) by LS
  .= a|^(n+1) + Sum(b*((a,b)Subnomial n)) by RVSUM_2:33
  .= b*Sum((a,b) Subnomial n) + a|^(n+1) by RVSUM_2:38;
  hence thesis by A1;
end;
