The nth term is (2n-1)(2n+1)=4n^2-1, where n starts at 1. The sum, S(n)=sum(4n^2-1)=4sum(n^2)-n.
The formula for the sum of the natural numbers is n^3/3+n^2/2+n/6, S(n)=4(n^3/3+n^2/2+n/6)-n.
Put n=4 to check it out: 4(64/3+8+2/3)-4=4*30-4=116=3+15+35+63=116.
S(50)=4(125000/3+2500/2+25/3)-50=4(125025)/3+1250)-50=4(41675+1250)-50=171,650.
Derivation of sum of natural squares
1+4+9+16+25
1st diff: 3 5 7 9
2nd diff: 2 2 2 (constant)
The sum can be generated by a cubic: An^3+Bn^2+Cn+D, where A, B, C and D are coefficients we need to find.
Put n=1: A+B+C+D=1 (1^2=1);
Put n=2: 8A+4B+2C+D=5 (1^2+2^2); subtract previous equation: 7A+3B+C=4;
Put n=3: 27A+9B+3C+D=14 (1^2+2^2+3^2);
Put n=4: 64A+16B+4C+D=30.
We now have 4 equations and 4 unknowns. Subtract eqn3 from eqn4: 37A+7B+C=16.
From this subtract 7A+3B+C: 30A+4B=12, so 15A+2B=6, B=(6-15A)/2. Substitute for B in equation involving A, B and C only: 7A+3(6-15A)/2+C=4, 7A+(18-45A)/2+C=4, 14A+18-45A+2C=8, 2C=31A-10, C=(31A-10)/2.
Now we have B and C in terms of A. D=1-(A+B+C)=1-A-(6-15A)/2-(31A-10)/2=(2-2A-6+15A-31A+10)/2=(6-18A)/2=3-9A. Now we have B, C, D in terms of A.
When we find A, we can find the other unknowns, so we need an equation we haven't used containing all 4 unknowns. Try the third equation:
27A+9(6-15A)/2+3(31A-10)/2+3-9A=14, 27A+(54-135A)/2+(93A-30)/2+3-9A=14, 54A+54-135A+93A-30+6-18A=28, -6A=-2, A=1/3. So B=(6-5)/2=1/2, C=(31/3-10)/2=1/6, D=0. So the formula is: n^3/3+n^2/2+n/6.