Combinatorial identity's algebraic proof without induction.

Question

How would you prove this combinatorial idenetity algebraically without induction?

$$\sum_{k=0}^n { x+k \choose k} = { x+n+1\choose n }$$

Thanks.

Answer 1

Here is an algebraic approach. In order to do so it's convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} \binom{n}{k}=[z^k](1+z)^n \end{align*}

We obtain

\begin{align*} \sum_{k=0}^{n}\binom{x+k}{k}&=\sum_{k=0}^{n}\binom{-x-1}{k}(-1)^k\tag{1}\\ &=\sum_{k=0}^n[z^k](1+z)^{-x-1}(-1)^k \tag{2}\\ &=[z^0]\frac{1}{(1+z)^{x+1}}\sum_{k=0}^n\left(-\frac{1}{ z }\right)^k\tag{3}\\ &=[z^0]\frac{1}{(1+z)^{x+1}}\cdot \frac{1-\left(-\frac{1}{z}\right)^{n+1}}{1-\left(-\frac{1}{z}\right)}\tag{4}\\ &=[z^n]\frac{z^{n+1}+(-1)^n}{(1+z)^{x+2}}\tag{5}\\ &=(-1)^n[z^n]\sum_{k=0}^\infty\binom{-x-2}{k}z^k\tag{6}\\ &=(-1)^n\binom{-x-2}{n}\tag{7}\\ &=\binom{x+n+1}{n}\tag{8} \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (2) we apply the coefficient of operator.

• In (3) we do some rearrangements by using the linearity of the coefficient of operator and we also use the rule \begin{align*} [z^{p-q}]A(z)=[z^p]z^{q}A(z) \end{align*}

• In (4) apply the formula for the finite geometric series.

• In (5) we do some simplifications and use again the rule stated in comment (3).

• In (6) we use the geometric series expansion of $\frac{1}{(1+z)^{x+2}}$. Note that we can ignore the summand $z^{n+1}$ in the numerator since it has no contribution to the coefficient of $z^n$.

• In (7) we select the coefficient of $z^n$.

• In (8) we use the rule stated in comment (1) again.

Answer 2

Hint: The R.H.S counts how many ways you can arrange $x+1$ $A$s and $n$ $B$s in a $x+n+1$ long sequence. As for the L.H.S, consider how many of these such sequences end in exactly $j$ $B$s as $j$ varies from $0$ to $n$. Hope this helps! Feel free to ask for extra clarification.

Answer 3

Suppose we seek to verify that

$$\sum_{k=0}^n {q+k\choose k} = {q+n+1\choose n}.$$

The difficulty here lies in the fact that the binomial coefficients on the LHS do not have an upper bound for the sum wired into them. We use an Iverson bracket to get around this:

$$[[0\le k\le n]] = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{w^k}{w^{n+1}} \frac{1}{1-w} \; dw.$$

Introduce furthermore

$${q+k\choose k} = {q+k\choose q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q+k} \; dz.$$

With the Iverson bracket in place we can let the sum range to infinity, getting

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w} \sum_{k\ge 0} w^k (1+z)^k \; dw\; dz.$$

This converges when $|w(1+z)| < 1.$ Simplifying we have

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w} \frac{1}{1-w(1+z)} \; dw\; dz.$$

Now the residues of the inner integral sum to zero so we can evaluate it by computing the negative of the residues at $w=1$ and at $w=1/(1+z).$ We get for the first one

$$- \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+2}} (1+z)^{q} \; dz = - [z^{q+1}] (1+z)^q = 0.$$

For the second one we have

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q-1} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w} \frac{1}{1/(1+z)-w} \; dw\; dz.$$

This yields

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q-1} (1+z)^{n+1} \frac{1}{1-1/(z+1)} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+2}} (1+z)^{q+n+1} \; dz = {q+n+1\choose q+1} = {q+n+1\choose n}.$$

This concludes the argument.

In order to be rigorous we must show that the residue at infinity of the inner integral is zero. We get

$$\mathrm{Res}_{w=\infty} \frac{1}{w^{n+1}} \frac{1}{1-w} \frac{1}{1-w(1+z)} \\ = - \mathrm{Res}_{w=0} \frac{1}{w^2} w^{n+1} \frac{1}{1-1/w} \frac{1}{1-(1+z)/w} \\ = - \mathrm{Res}_{w=0} w^{n+1} \frac{1}{w-1} \frac{1}{w-(1+z)}.$$

There is most certainly no pole here an the residue is zero as claimed (note that $1+z$ circles the value one.)

Addendum. I just realized that it is actually a lot simpler without the Iverson bracket. We get

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q} \sum_{k=0}^n (1+z)^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} (1+z)^{q} \frac{(1+z)^{n+1}-1}{1+z-1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+2}} (1+z)^{q} ((1+z)^{n+1}-1) \; dz.$$

This has two pieces, the second is

$$-[z^{q+1}] (1+z)^q = 0$$

and the first is

$$[z^{q+1}] (1+z)^{q+n+1} = {q+n+1\choose q+1} = {q+n+1\choose n}.$$