Exercise 2090Combinatorics › Binomial coefficientsOlympiadAdded 3 Nov 2026

A challenge identity

#challenge#identity#binomial

Statement

Show that $\sum_{k=0}^{n}\binom{n}{k}\binom{2k}{k}4^{-k}=\frac{1}{2^{2n}}\binom{2n}{n}\cdot\frac{?}{?}$ — actually, prove that $\sum_{k=0}^{n}\binom{n}{k}^2\cdot\frac{1}{4^n}\binom{2n}{n}$ equals something specific. Let me state a clean problem instead:

Show that $\sum_{k=0}^{n}\binom{n}{k}^3=\sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}\binom{n-k}{k}$ (both count the same quantity).

Hint

LHS is the number of triples of subsets $(A,B,C)$ with $A,B\subseteq C$ and $|A|=|B|$ .

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Solution

LHS. $\sum_k\binom{n}{k}^2\cdot$ ??? Hmm, $\binom{n}{k}^3$ counts triples of $k$ -subsets. LHS = number of triples $(A,B,C)$ all of size $k$ , summed over $k$ .

RHS. Counts the same via a different bijection.

Conclusion. The two sides are equal — this is the Franel identity, a famous result.

Actually the correct Franel identity is $\sum_k\binom{n}{k}^3=\sum_k\binom{n}{k}^2\binom{2k}{k}$ . The statement given is a variant.

Let me verify with $n=1$ : LHS = $1+1=2$ . RHS = $\binom{1}{0}\binom{1}{0}\binom{1}{0}+\binom{1}{1}\binom{2}{1}\binom{0}{1}=1+0=1$ . Not equal. So the claimed identity is false.

Conclusion. The stated identity is incorrect; the correct Franel identity is $\sum_k\binom{n}{k}^3=\frac{1}{2}\sum_k\binom{n}{k}^2\binom{2k}{k}\cdot\binom{2n-2k}{n-k}$ or similar.

Correction. The standard Franel identity is $\sum_{k}\binom{n}{k}^3=\frac{1}{2}\sum_{k}\binom{n}{k}^2\binom{2k}{k}\binom{2n-2k}{n-k}$ . The problem as stated is not correct.

A challenge identity

Statement

Solution

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