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If someone could provide some explanation/proof, it would be greatly appreciated.

$$\binom{n+2}{3} = \sum_{i=1}^{n} i(n-i+1)$$

Q: What's the reasoning behind the RHS?

Shaun
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bli00
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    Your second question has been answered here. – N. F. Taussig Jan 29 '18 at 18:22
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    First part- Need to select $3$ distinct numbers out of $1,..,n+2$. Any $3$ numbers will have a middle element. So we can choose a middle element and then proceed to choose the smaller and bigger elements. Number of $3$-tuples with $i+1$ as the middle element will be $i(n-i+1)$. Sum this over $i$. – Manan Jan 29 '18 at 18:24
  • @N. F. Taussig fixed the question, thank you. – bli00 Jan 29 '18 at 19:40

2 Answers2

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For the first question count the three element subsets of $\{0,\dotsc,n+1\}$ in two different ways. There are $\binom{n+2}{3}$ such subsets. Classify the subsets based on their middle element (i.e. the element which is not maximum and not minimum). If the middle element is $i$, then there are $i$ choices for the minimum element and $(n+1)-(i+1)+1=n-i+1$ choices for the maximum element where $1\leq i\leq n$.

Sri-Amirthan Theivendran
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Using the variation of Vandermonde's Identity proven in this answer, $$ \begin{align} \sum_{i=1}^ni(n-i+1) &=\sum_{i=0}^{n+1}\binom{i}{1}\binom{n+1-i}{1}\\ &=\binom{n+2}{3} \end{align} $$

robjohn
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