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Problem: Prove combinatorically that $ \sum_{k=1}^{n} k^2 = n{n+1 \choose 2} - {n+1 \choose 3 } $.

My thoughts: I rewrote the equation to be proved as $ \sum_{k=1}^{n} k^2 = {n \choose 1 }{n+1 \choose 2} - { n \choose 0 }{n+1 \choose 3 } $
Then I was thinking to myself that $ k^2 $ on the left side could represent the number of length-2 strings over alphabet of size $ k$.
And the right hand side could be interpreted as: Let $ A $ represent the set of $ 2n+1 $ people composed of two sets: $ B $ the set of $ n+1 $ boys and $ C $ the set of $ n $ girls. $ {n \choose 1 }{n+1 \choose 2} $ could be thought as the set of all possible ways to create a commission made from $ 2 $ boys and $ 1 $ girl ( and similarly for $ { n \choose 0 }{n+1 \choose 3 } $ ).

However, thinking of $ A $ as made from sets $ B ,C $ of boys and girls doesn't seem to make sense of the difference in the right side between the binomials. Also, thinking of the left hand side as a way of choosing number of length-2 strings over alphabet of size $ k$ doesn't make me see how the left-side correlates to the right-hand side. Can you please help me on how to prove the equation?

hazelnut_116
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    This problem has been discussed at some length here: https://math.stackexchange.com/questions/95047/combinatorial-interpretation-of-sum-of-squares-cubes – Vectorizer Jun 04 '21 at 10:10
  • Thanks, I'll read it. I went over some posts before posting but didn't notice this. I still have to prove that $ \sum_{k=1}^n k^2 = \binom{n+1}{2} + 2 \binom{n+1}{3} = n{n+1 \choose 2} - {n+1 \choose 3 } $ – hazelnut_116 Jun 04 '21 at 10:14
  • Ok, I understand when the problem is phrased as $ \binom{n+1}{2} + 2 \binom{n+1}{3} $. But I have difficulty for the form $ n{n+1 \choose 2} - {n+1 \choose 3 } $. I thought about it, we are creating a triple $ (i,j,k) $. For $ n{n+1 \choose 2} $, $ {n+1 \choose 2} $ means we choose $ 2 $ elements from the set $ {0,...,n } $ and $ n $ means we chose an element from the set ${1,...,n }$ , we'll have 3 elements in total, we don't care about two elements but we take some third element in these choices to be the largest or equal from those two. But why do I have " $- {n+1 \choose 3 }$ " ? – hazelnut_116 Jun 04 '21 at 10:45
  • The triple $ (i,j,k) $ is such that $ 0 \leq i \leq j \leq k-1$ and $ 1\leq k \leq n $ – hazelnut_116 Jun 04 '21 at 10:46

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The boy is numbered from $1$ to $n$, the girls from $1$ to $n+1$. We count the number of triples $(leftgirl,boy,rightgirl)$ with $boy$ less than both $girl$s. For example $(2,1,3)$ or $(6,5,6)$.
The boy has a girl's photo in each hand, possibly the same one, and the girls both have a higher number than the boy. Boy $1$ has $n$ choices for each photo, but boy $n$ only has one choice for each photo. That is the sum of $k^2$.
The boy and the left-hand girl can be any pair of numbers, with the girl the higher number. That is $n+1\choose 2$. The right-hand girl can be any number not equal to the boy. That is $n$.
But now remove those with the right-hand girl less than the boy. That is $n+1\choose3$.

Empy2
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  • I didn't fully understand what is the set you are counting - is it the set of pairs of boys and girls ranked? ( like $ \langle boy_1, girl_2 \rangle $ where $boy_1$<$ girl_2 $ ) can you please give me an example of an element of the set? – hazelnut_116 Jun 04 '21 at 11:35