let s be Real_Sequence; :: thesis: ( ( for n being Element of NAT st n >= 1 holds
( s . n = (2 |^ n) * (((3 * n) - 1) / 4) & s . 0 = 0 ) ) implies for n being Element of NAT st n >= 1 holds
(Partial_Sums s) . n = ((2 |^ n) * (((3 * n) - 4) / 2)) + 2 )
assume A1: for n being Element of NAT st n >= 1 holds
( s . n = (2 |^ n) * (((3 * n) - 1) / 4) & s . 0 = 0 ) ; :: thesis: for n being Element of NAT st n >= 1 holds
(Partial_Sums s) . n = ((2 |^ n) * (((3 * n) - 4) / 2)) + 2
defpred S₁[ Nat] means (Partial_Sums s) . $1 = ((2 |^ $1) * (((3 * $1) - 4) / 2)) + 2;
(Partial_Sums s) . (1 + 0 ) = ((Partial_Sums s) . 0 ) + (s . (1 + 0 )) by SERIES_1:def 1
.= (s . 0 ) + (s . 1) by SERIES_1:def 1
.= 0 + (s . 1) by A1
.= (2 |^ 1) * (((3 * 1) - 1) / 4) by A1
.= 2 * (1 / 2) by NEWTON:10
.= (((3 - 4) / 2) * 2) + 2
.= (((3 - 4) / 2) * (2 |^ 1)) + 2 by NEWTON:10
.= (((2 |^ 1) * ((3 * 1) - 4)) / 2) + 2 ;
then A2: S₁[1] ;
A3: for n being Nat st n >= 1 & S₁[n] holds
S₁[n + 1]

proof

let n be Nat; :: thesis: ( n >= 1 & S₁[n] implies S₁[n + 1] )
assume A4: ( n >= 1 & (Partial_Sums s) . n = ((2 |^ n) * (((3 * n) - 4) / 2)) + 2 ) ; :: thesis: S₁[n + 1]
A5: n + 1 >= 1 by NAT_1:11;
n in NAT by ORDINAL1:def 13;
then (Partial_Sums s) . (n + 1) = (((2 |^ n) * (((3 * n) - 4) / 2)) + 2) + (s . (n + 1)) by A4, SERIES_1:def 1
.= (((2 |^ n) * (((3 * n) - 4) / 2)) + 2) + ((2 |^ (n + 1)) * (((3 * (n + 1)) - 1) / 4)) by A1, A5
.= (((2 |^ n) * (((3 * n) - 4) / 2)) + ((2 |^ (n + 1)) * (((3 * n) + 2) / 4))) + 2
.= (((2 |^ n) * (((3 * n) - 4) / 2)) + (((2 |^ n) * 2) * (((3 * n) + 2) / 4))) + 2 by NEWTON:11
.= (((2 |^ n) * 2) * (((3 * n) - 1) / 2)) + 2
.= ((2 |^ (n + 1)) * (((3 * (n + 1)) - 4) / 2)) + 2 by NEWTON:11 ;
hence S₁[n + 1] ; :: thesis: verum

end;

for n being Nat st n >= 1 holds
S₁[n] from NAT_1:sch 8(A2, A3);
hence for n being Element of NAT st n >= 1 holds
(Partial_Sums s) . n = ((2 |^ n) * (((3 * n) - 4) / 2)) + 2 ; :: thesis: verum