#1**+1 **

Maybe easier ways to do this......but.....using the sum of differences we have

1 5 14 30 55

4 9 16 25

5 7 9

2 2

We have 3 non-zero results so we can express the progessive sums as a cubic polynomial

an^3 + bn^2 + cn + d where n is the sum of the first n squares

We have that

a(1)^3 + b(1)^2 + c(1) + d = 1

a(2)^3 + b(2)^2 + c(2) + d = 5

a(3)^3 + b(3)^2 + c(3) + d = 14

a(4)^3 + b(4)^2 + c(4) + d = 30

a + b + c + d = 1

8a + 4b + 2c+ d = 5

27a + 9b + 3c + d = 14

64a + 16b + 4c + d = 30

The solution to this is

a = 1/3 b = 1/2 c =1/6 d = 0

So.....the resulting polynomial is

(1/3)n^3 + (1/2)n^2 + (1/6)n

So.....the sum of the first 1000 squares is

(1/3) (1000)^3 + (1/2)(1000)^2 + (1/6)(1000) =

333,833,500

CPhill Feb 10, 2020