The summation, $$\sum_{i=1}i^2=n(n+1)(2n+1)/6$$ However, how could you prove this? All of the proofs I've seen already assume knowledge of the formula, but how do you prove this without first knowing the formula? What about formulas for higher powers, such as cubics and quartics? If possible, please keep this on a level where I can understand (I'm in calculus AB). Thanks!
Answer
Polya gives a beautiful discussion of this problem in Chapter 3 (called "Recursion") in volume 1 of his book "Mathematical Discovery." One way to "discover" the formula is to exploit the identity $$(n+1)^3=n^3+3n^2+3n+1$$ which implies $$(n+1)^3-n^3=3n^2+3n+1$$
If you sum all such equations from $n=1$ to $n=k$ you find the left side telescopes, leaving $$(k+1)^3-1=3S+3(1+2+\cdots+k)+k$$
where $S$ is the sum you wanted. Of course, the sum in parentheses is just $\frac{k(k+1)}{2}$, and solving the equation for $S$ after making that substitution gives the result.
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