reserve k, l, m, n, i, j for Nat,
  K, N for non empty Subset of NAT,
  Ke, Ne, Me for Subset of NAT,
  X,Y for set;
reserve f for Function of Segm n,Segm k;
reserve x,y for set;

theorem Th48:
  n >= 2 implies n block 3 = 1/6 * ( 3 |^ n - 3 * 2 |^ n + 3 )
proof
  defpred P[Nat] means $1 block 3=1/6*(3|^$1-3*2|^$1+3);
A1: for k be Nat st k>=2 & P[k] holds P[k+1]
  proof
    let k be Nat such that
A2: k>=2 and
A3: P[k];
    k block 2 = 1/2 * (2 |^ k - 2) by A2,Th47,XXREAL_0:2;
    hence (k+1) block 3 = 3*(k block (2+1)) +(1/2*(2|^k-2)) by Th46
      .=1/6*((3*3|^k)-3*2*2|^k +3) by A3
      .=1/6*(3|^(k+1)-3*(2*2|^k) +3) by NEWTON:6
      .=1/6*(3|^(k+1)-3*2|^(k+1) +3) by NEWTON:6;
  end;
  1/6*(3|^2-3*2|^2+3)=1/6*(3*3-3*2|^2+3) by WSIERP_1:1
    .=1/6*(3*3-3*(2*2)+3) by WSIERP_1:1
    .=2 block 3 by Th29;
  then
A4: P[2];
  for k be Nat st k>=2 holds P[k] from NAT_1:sch 8(A4,A1);
  hence thesis;
end;
