The absorption identity for binomial coefficients is This post will give a couple of proofs of the identity and then use it to prove some other identities involving binomial coefficient sums.

**Proofs of the absorption identity**

*Proof 1: Use the factorial definition of the binomial coefficients.*

*Proof 2: A combinatorial argument.*

Combinatorially, the absorption identity is probably better viewed in the form

In this reformulation, both sides count the number of ways to choose a chaired committee of size from total people. The left-hand side counts the number of ways to choose the committee members first and then choose a chair from those people. The right-hand side counts the number of ways to choose the chair first from people and then choose the remaining committee members from the remaining people.

**The first summation identity (a two for one, really)**

Let’s find the value of

The absorption identity works beautifully here, as it allows us to replace the factor of in the denominator with a factor of that does not depend on the summation index :

since the alternating sum of a row of binomial coefficients is .

(At this math.SE question André Nicolas gives a different proof using integration and the binomial formula, and I give a probabilistic proof that uses inclusion-exclusion.)

A similar argument shows that (See, for example, my answer here.)

**The second summation identity **

Let’s find the value of the similar sum

We can’t use the absorption identity directly here because we have in the denominator of the summand rather than . However, by combining finite differences and the recursion formula with the absorption identity we can find the sum.

Let . Then

Therefore,

where is the *n*th harmonic number.

**A bonus summation identity**

Applying the absorption identity twice with the ideas in the derivation of the previous identity shows that

For the full derivation, see my answer here.

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