How do you prove this

Ok so I know that 1^2 + 2^2 + 3^2 + 4^2..........+ n^2 = [n(n+1)(2n+1)]/6
Can someone prove this to me? One of the websites which I checked did it with "mathematical induction". I dont know anything about it. I study in class 10 and want to know another way for proof.
You can use concepts related to polynomials, arithmetic progressions, geometric progressions and others taught till class 10. I dont know about calculus either.
Sorry for the restrictions. Thank you.

Let be S the sum
(1+0)^3 = 1^3
(1+1)^3 = 1^3 + 3*1^2*1^1 + 3*1^1*1^2 + 1^3 = 2^3
(1+2)^3 = 1^3 + 3*1^2*2^1 + 3*1^1*2^2 + 2^3 = 3^3
- - - - - -
(1+n)^3 = n^3 + 3n + 3n^2 + 1
......................................… add them
1^3 + . . + (1+n)^3 = (n+1) + 3(1+2 + . . . n) + 3(1^2 + 2^2 + . . n^2) + (1^3 + ... n^3)
(1+n)^3 = (n+1) + 3n(n+1)/2 + 3S
S = (1/3)[ (1+n)^3 - (n+1) - 3n(n+1)/2 ] =
(1/6)[2n^3 + 6n^2 + 6n + 2 - 2n - 2 - 3n^2 - 3n] =
(1/6)[2n^3 + 3n^2 + n ] =
(1/6)n(2n^2 + 3n + 1) =
(1/6)n(n+1)(2n + 1)

This answer probably won't be the neatest way to answer this problem but it may answer any questions you have such as "How did anyone figure out this formula?"

You don't need to use any intuition to find the formula for the above sum, in fact it can all be done methodically using differences. Any function which eventually has a "divided difference" another polynomial is a polynomial itself.

Let p(n) be an expression for the above equation

Then p(n+1) - p(n) = n^2 + 2n + 1.
The second difference is (n+1)^2 + 2(n+1) + 1 - n^2 - 2n - 1 = 2n + 3
The third difference is 2. We divide by 3*2 to find the leading coefficient = 1/3

(n+1)^3 / 3 - n^3/3 = n^2 + n + 1/3. So if p(n+1) - p(n) = n^2 + 2n + 3, the rest of p(n) must have difference n + 8/3. Giving a second difference of 1. Thus the quadratic coefficient is 1/2.

Do this again and we notice the linear term is 1/6 and there is no constant term.