reserve a,b,c for set;
reserve r for Real,
  X for set,
  n for Element of NAT;

theorem Th22:
  0 < r & r < 1 implies Sum ((r GeoSeq)^\n) = r|^n / (1-r)
proof
  assume that
A1: 0 < r and
A2: r < 1;
A3: |.r.| = r by A1,ABSVALUE:def 1;
  then
A4: r GeoSeq is summable by A2,SERIES_1:24;
A5: dom ((r|^n)(#)(r GeoSeq)) = NAT by FUNCT_2:def 1;
  now
    let i being Element of NAT;
    thus ((r GeoSeq)^\n).i = (r GeoSeq).(i+n) by NAT_1:def 3
      .= r|^(i+n) by PREPOWER:def 1
      .= (r|^n)*(r|^i) by NEWTON:8
      .= (r|^n)*((r GeoSeq).i) by PREPOWER:def 1
      .= ((r|^n)(#)(r GeoSeq)).i by A5,VALUED_1:def 5;
  end;
  then
A6: (r GeoSeq)^\n = (r|^n)(#)(r GeoSeq) by FUNCT_2:63;
  Sum (r GeoSeq) = 1/(1-r) by A3,A2,SERIES_1:24;
  hence Sum ((r GeoSeq)^\n) = r|^n *(1/(1-r)) by A4,A6,SERIES_1:10
    .= (r|^n*1) / (1-r) by XCMPLX_1:74
    .= (r|^n) / (1-r);
end;
