I noticed that the difference between two consecutive sums of N (integer) consecutive squares is equal to 2 x N. Is this trivial or already known and generalisable for all N?
Not sure if this is correct notation, but
Let:
$$ A = \sum_{N=1}^{N} ((i+2)^2 + (i+3)^2 + (i+4)^2 ...+i_N^{2}) - \sum_{N=1}^{N} ((i+1)^2 + (i+2)^2 + (i+3)^2 + ...+i_N^{2})$$
$$ B = \sum_{N=1}^{N} ((i+1)^2 + (i+2)^2 + (i+3)^2 ...+i_N^{2}) - \sum_{N=1}^{N} (i^2 + (i+1)^2 + (i+2)^2 + ...+i_N^{2}) $$ Then $$A - B = 2N $$
For example (N=3): $$X =(2^2 + 3^2 + 4^2) -(1^2 + 2^2 + 3^2) = 29-14 = 15 $$ $$Y = (3^2 + 4^2 + 5^2) -(2^2 + 3^2 + 4^2) = 50-29 = 21 $$ $$Z = (4^2 + 5^2 + 6^2) -(3^2 + 4^2 + 5^2) = 77-50 = 27 $$
$$Z-Y = 27 - 21 = Y-X = 21 - 15 = 2N = 2(3) = 6$$
For example (N=9): $$X =(2^2 + 3^2 + ... + 10^2) - (1^2 + 2^2 + ... + 9^2) = 384-285 = 99 $$ $$Y = (3^2 + 4^2 + ... + 11^2) - (2^2 + 3^2 + ... + 10^2) = 501-384 = 117 $$ $$Z = (4^2 + 5^2 + ... + 12^2) - (3^2 + 4^2 + ... + 11^2) = 636-501 = 135 $$
$$Z-Y = 135 - 117 = Y-X = 117 - 99 = 2N = 2(9) = 18 $$
This seems to generalise for all consecutive sums of N consecutive squares. I noticed it when learning about Legendres Three Square Theorem and Lagranges Four Square Theorem
Denote $$S(a,b):=a^2+(a+1)^2+\cdots +(b-1)^2+b^2=\sum_{j=a}^b j^2$$
For a given $N$, you have $$X=S(2,N+1)-S(1,N)=(N+1)^2-1=N^2+2N$$
$$Y=S(3,N+2)-S(2,N+1)=(N+2)^2-4=N^2+4N$$
$$Z=S(4,N+3)-S(3,N+2)=(N+3)^2-9=N^2+6N$$
Therefore , we have $Z-Y=Y-X=2N$