'Oddy Squary' Sum!

First we will find the sum for $n$ terms.The given sum can be written as:

$\large\displaystyle\sum_{i=1}^n (2i-1)^2 = \displaystyle\sum_{i=1}^n 4i^2 - 4i + 1 = 4\displaystyle\sum_{i=1}^n i^2 -4\displaystyle\sum_{i=1}^ni + \sum_{i=1}^n 1$

$=4\dfrac{n(n+1)(2n+1)}{6} - 4\dfrac{n(n+1)}{2} + n$

$=4\left(\dfrac{n(n+1)}{2}\right)\left(\dfrac{2n+1}{3} - 1\right) + n$

$=\dfrac{4n(n+1)(n -1)}{3} + n$

$=\dfrac{4n(n^2 -1)+3n}{3}$

$=\dfrac{n(4n^2 - 1)}{3}$

Substituting $n = 500$ , $S =166666500$

Thus , $\dfrac{S}{333} = \dfrac{166666500}{333} = \huge\color{#D61F06}{\boxed{500500}}$

Firstly, I must state an important equation: $\displaystyle\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \ (1)$ Now I will rewrite the equation: $1^2 + 3^2 +5^2 +...+999^2 = (1^2 +2^2 +...+1000^2) - (2^2 +4^2+...+1000^2)$ $= (1^2 +2^2 +...+1000^2) - 4(1^2 +2^2 +...+500^2)\ using \ (1).....$ $= \frac{1000 \times\ 1001 \times\ 2001}{6} - 4 (\frac{500 \times\ 501 \times\ 1001}{6}) = 166666500 = S$ $\huge\color{#20A900}{\therefore\frac{S}{333} = \boxed{500500}}$

This sum can be rewritten as :

1 + 9 + 25 … + 998001 = S

It’s a second-degree arithmetical progression therefore can be written as a third-degree polinomy:

S(n) = an^3 + bn^2 + cn + d

S(1) = 1 ; S(3) = 10 ; S(5) = 35 ; S(7) = 84

I – for n = 1: a + b + c + d = 1

II – for n=3: 27a + 9b + 3c + d = 10

III – for n=5: 125a + 25b + 5c + d = 35

IV – for n=7: 343a + 49b + 7c + d = 84

Solving the system: a = 1/6 ; b = 1/2 ; c = 1/3 ; d = 0

S(n) = (n^3 + 3n^2 + 2n)/6

For n=999 : S = 166666500

For last: S/333 = 500500

2n-1=999 so, n=500 s=(n(2n+1)(2n-1))/3 =1666666500 so, S/333= 500500