How to Prove Square Pyramidal Number Formula

Statement

∑nk=1k^2=n(n+1)(2n+1)/6

Proof

1^2=1

2^2=4=1+(2*1+1)

3^2=9=(2*1+1)+(2*3+1)

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(n-1)^2=1+(2*1+1)+(2*3+1)+...+(2(n-2)+1)

n^2=1+(2*1+1)+(2*3+1)+...+(2(n-2)+1)+(2(n-1)+1)

add them all together, and we can get the following equality

∑nk=1k^2=1n+(2*1+1)(n-1)+...+(2(n-2)+1)*2+(2(n-1)+1)*1

=∑nk=1(2(k-1)+1)(n-k+1)=∑nk=1(2k-1)(n-k+1)

→3∑nk=1k^2=n(2n+3)(n+1)/2-n^2-n

→∑nk=1k^2=n(n+1)(2n+1)/6

Q.E.D.

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