1 squared plus 2 squared .... What is the sum of squares that has been added to n? is there a formula?

Square sum formula N (n+1) (2n+1)/6
i.e. 1^2+2^2+3^2+...+n^2=n (n+1) (2n+1)/6 (note: square of N^2=n)
Prove 1+4+9+ ... +n^2=n (n+1) (2n+1)/6
Evidence Law One (inductive conjecture method):
1, N=1, 1=1 (2x1+1)/6=1
2, n=2, 1+4=2 (2+1) (2x2+1)/6=5
3, set n=x, the formula is set up, namely 1+4+9+ ... +x2=x (x+1) (2x+1)/6
When the n=x+1 is
1+4+9+ ... +x2+ (x+1) 2=x (x+1) (2x+1)/6+ (x+1) 2
= (x+1) [2 (x2) +x+6 (x+1)]/6
= (x+1) [2 (x2) +7X+6]/6
= (x+1) (2x+3) (x+2)/6
= (x+1) [(x+1) +1][2 (x+1) +1]/6
Also satisfies the formula
4, in summary, the square and Formula 1^2+2^2+3^2+...+n^2=n (n+1) (2n+1)/6 was established, the certificate.
Certificate Law II (using identity (n+1) ^3=n^3+3n^2+3n+1):
(n+1) ^3-n^3=3n^2+3n+1,
n^3-(n-1) ^3=3 (n-1) ^2+3 (n-1) +1
.
3^3-2^3=3* (2^2) +3*2+1
2^3-1^3=3* (1^2) +3*1+1.
Add the two ends of the n equation separately to:
(n+1) ^3-1=3 (1^2+2^2+3^2+.+n^2) +3 (1+2+3+...+n) +n,
Due to 1+2+3+...+n= (n+1) N/2,
On behalf of the people to:
N^3+3n^2+3n=3 (1^2+2^2+3^2+.+n^2) +3 (n+1) n/2+n
After finishing:
1^2+2^2+3^2+.+n^2=n (n+1) (2n+1)/6