let s be Real_Sequence; :: thesis: ( ( for n being Nat holds s . n = n |^ 2 ) implies for n being Nat holds (Partial_Sums s) . n = ((n * (n + 1)) * ((2 * n) + 1)) / 6 )
defpred S₁[ Nat] means (Partial_Sums s) . $1 = (($1 * ($1 + 1)) * ((2 * $1) + 1)) / 6;
assume A1: for n being Nat holds s . n = n |^ 2 ; :: thesis: for n being Nat holds (Partial_Sums s) . n = ((n * (n + 1)) * ((2 * n) + 1)) / 6
A2: for n being Nat st S₁[n] holds
S₁[n + 1]

proof

let n be Nat; :: thesis: ( S₁[n] implies S₁[n + 1] )
assume (Partial_Sums s) . n = ((n * (n + 1)) * ((2 * n) + 1)) / 6 ; :: thesis: S₁[n + 1]
then (Partial_Sums s) . (n + 1) = (((n * (n + 1)) * ((2 * n) + 1)) / 6) + (s . (n + 1)) by SERIES_1:def 1
.= (((n * (n + 1)) * ((2 * n) + 1)) / 6) + ((n + 1) |^ 2) by A1
.= (((n * (n + 1)) * ((2 * n) + 1)) + (((n + 1) |^ 2) * 6)) / 6
.= ((((n + 1) * n) * ((2 * n) + 1)) + (((n + 1) * (n + 1)) * 6)) / 6 by WSIERP_1:1 ;
hence S₁[n + 1] ; :: thesis: verum

end;

(Partial_Sums s) . 0 = s . 0 by SERIES_1:def 1
.= 0 |^ 2 by A1
.= ((0 * (0 + 1)) * ((2 * 0) + 1)) / 6 by NEWTON:11 ;
then A3: S₁[ 0 ] ;
for n being Nat holds S₁[n] from NAT_1:sch 2(A3, A2);
hence for n being Nat holds (Partial_Sums s) . n = ((n * (n + 1)) * ((2 * n) + 1)) / 6 ; :: thesis: verum