How to prove that $1^2+2^2+\cdot\cdot\cdot+n^2=\frac{n(n+1)(2n+1)}{6}$ with a combinatorial proof

Question

Prove that $1^2+2^2+\cdot\cdot\cdot+n^2=\frac{n(n+1)(2n+1)}{6}$ by using only Double Counting.

(Hint: Count the triples of $(x,y,z)$ for some conditions)

I can only prove the left side.

Can you show your work? You said "I can only prove the left side". — A. P., Apr 29 '22 at 13:22
You should, at the very least, give the conditions on the triples that allow you to establish the LHS. — Brian Tung, Apr 29 '22 at 15:49

score 1 · Accepted Answer · answered Apr 29 '22 at 16:11

1

It is a very elegant (but long) proof which deals with counting the elements of the set $$T_n=\{(a,b,c)\in S_n:a<c,\,b<c\}$$ where $$S_n=\{1,2,3,\dots ,n,n+1\}$$ The idea is to first fix $c$ and choose $a$ and $b$ keeping the conditions in mind. This gives the expression $$\sum_{k=1}^n k^2$$ And then try to use the fact that in a triple, either $a=b$ or $a\neq b$. this will give you the RHS.

You can find the details here.

answered Apr 29 '22 at 16:11

Sayan Dutta

8,831

1

Thank you. In fact, they gave this question with 1 page to write the solution. – Jiras K. Apr 30 '22 at 14:31

How to prove that $1^2+2^2+\cdot\cdot\cdot+n^2=\frac{n(n+1)(2n+1)}{6}$ with a combinatorial proof

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