<s>[INST] Seeking non-inductive, combinatorial proof of the identity $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ How do you prove 

$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$$

without induction?
I'm looking for a combinatorial proof of this. 
 [/INST] We can count the sum $$2^2 + 4^2 + \dots + (2n)^2 = {2n + 2 \choose 3}$$ as follows:
Choose three numbers from $\{1, 2, \dots, 2n+2\}$. If the largest number chosen is odd, notate it $2m+1$ - the other two smaller numbers can be chosen in ${2m \choose 2}$ ways. If the largest number chosen is $2m+2$, the other two numbers can be chosen in ${2m+1 \choose 2}$ ways. From the identity
$${2m \choose 2} + {2m+1 \choose 2} = (2m)^2$$
and summing this for $m = 0, 1, 2, \dots, n$, the even result follows and the general result follows from dividing by two.
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