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How to prove following Identity? $$\sum_{k=0}^n (-1)^k {n-k \choose k} m^k (m+1)^{n-2k} = \frac {m^{n+1}-1}{m-1}, m \ge 2$$

This seems very hard to me. Any idea about how to prove it combinatorialy?

[P.S: for $k > \lfloor {\frac n 2}\rfloor$ L.H.S evaluates to $0$. ]

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  • $\begingroup$ i just changed the power of (m+1) in the L.H.S . That was a mistake made by me. $\endgroup$ – arindam mitra Oct 28 '14 at 10:15
  • $\begingroup$ Your terms in the sum will equal zero as soon as k $\geq \frac{n}{2} +1$, since k $ > n-k$ which implies that the ${n-k \choose k}$ will be null. That doesn't look too good :) $\endgroup$ – mvggz Oct 28 '14 at 10:30
  • $\begingroup$ @DigitalBrain What makes you think it's incorrect? It checks out for $n\le4$. $\endgroup$ – bof Oct 28 '14 at 11:53
  • $\begingroup$ @mvggz Of course $\binom xk=0$ for $x=0,1,2,\dots,k-1$. You can write the upper index of summation as $\lfloor n/2\rfloor$ if you think it looks better that way. It doesn't change anything. $\endgroup$ – bof Oct 28 '14 at 11:56
  • $\begingroup$ I didn't say it was false, I said that it was not all too pretty :) . But yes I think it'd be maybe a bit clearer if the summation was from 0 to $\lfloor \frac{n}{2}\rfloor$ $\endgroup$ – mvggz Oct 28 '14 at 12:00
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Here's a combinatorial proof based on rewriting the right hand side as $1+m+m^2+\cdots+m^n$ and using inclusion-exclusion for the left hand side.

Suppose you have jars labeled $1,2,3,\ldots,n$, each containing balls labeled $0,1,2,\ldots,m$. Starting with jar $1$, you pick one ball from each jar, but stopping as soon as you've picked a ball labeled $0$. Or more precisely, you stop making choices, and just systematically take the $0$ ball from all the remaining jars. The number of different ways you'll wind up with exactly the first $k$ balls not labeled $0$ (and the rest all labeled $0$) is $m^k$, so the right hand side counts the total number of different outcomes.

So does the left hand side, except it does so by way of inclusion-exclusion, which I'll describe in a somewhat leisurely fashion.

Start by taking a ball from each jar, ignoring the stopping rule. This gives $(m+1)^n$ outcomes. The ones we don't want are those that have a $0$ followed by a nonzero. So let's get rid of these by counting the number of ways we can pick two consecutive jars, take the $0$ from the first and pick a nonzero from the second, finally picking whatever we like from all the rest. It's obvious there are $n-1$ ways to pick two consecutive jars (the first can be any but jar $n$). There are $m$ choices for the nonzero ball taken from the second jar in the bundled pair, and $(m+1)$ choices for each of the other $n-2$ jars. We'll write the multiplicative total in the suggestive form ${n-1\choose1}m(m+1)^{n-2}$, using the binomial coefficient for the factor $n-1$. This is the count we subtract.

But of course this first round of exclusion counts (and excludes) some outcomes multiple times, so we need to put some back. In particular we need to restore outcomes in which there are two pairs of consecutive jars with a $0$ taken from the first and a nonzero from the second. A little stars-and-bars tells us there are ${n-2\choose2}$ ways to pick two (disjoint) pairs of consecutive jars. (The pairs have to be disjoint, since we're required to take the $0$ ball from the first jar in each pair and a nonzero ball from the second.) Having done so, there are $m^2$ choices for the nonzero balls in the two pairs and $(m+1)^{n-4}$ choices for the $n-4$ remaining jars. So we add back in the count ${n-2\choose2}m^2(m+1)^{n-4}$.

But this of course adds back too much. So the inclusion-exclusion continues. At the $k$th round, we need to pick $k$ disjoint pairs of consecutive jars, etc. The stars-and-bars argument tells us there are ${n-k\choose k}$ ways to make that set of choices, and the rest is $m^k$ followed by $(m+1)^{n-2k}$ as before, for a multiplicative total of ${n-k\choose k}m^k(m+1)^{n-2k}$. Putting these in- and ex-clusions all together, we get the left hand side.

(A final remark: The original version of this answer might be worth a look as a failed attempt to apply inclusion-exclusion. In it, I managed to assign a meaning to each term in the left hand side, but the inclusions and exclusions they represented were utterly meaningless. I had to sleep on the problem before I came up with a correct interpretation.)

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Here is another algebraic proof, somewhat involved where the calculations are concerned but simple conceptually.

We seek to compute $$\sum_{k=0}^n {n-k\choose k} (-1)^k m^k (m+1)^{n-2k}$$ or $$(m+1)^n \sum_{k=0}^n {n-k\choose k} (-1)^k m^k (m+1)^{-2k}.$$

Introduce the integral representation $${n-k\choose k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-k}}{z^{k+1}} \; dz.$$

This gives for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \sum_{k=0}^n (-1)^k m^k (m+1)^{-2k} (1+z)^{-k} z^{-k}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \sum_{k=0}^n \left(-\frac{m}{(m+1)^2 (1+z) z}\right)^k\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \frac{1-\left(-\frac{m}{(m+1)^2 (1+z) z}\right)^{n+1}} {1 + m/(m+1)^2/(1+z)/z}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z(1+z)} \frac{1-\left(-\frac{m}{(m+1)^2 (1+z) z}\right)^{n+1}} {1 + m/(m+1)^2/(1+z)/z}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} (1+z)^{n+1} \frac{1-\left(-\frac{m}{(m+1)^2 (1+z)z}\right)^{n+1}} {(1+z)z + m/(m+1)^2}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n+1}-\left(-\frac{m}{(m+1)^2 z}\right)^{n+1}} {(1+z)z + m/(m+1)^2}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{z^{n+1} (1+z)^{n+1}-\left(-\frac{m}{(m+1)^2}\right)^{n+1}} {(1+z)z + m/(m+1)^2}\; dz.$$

The first term in the numerator does not contribute and we are left with $$-\left(-\frac{m}{(m+1)^2}\right)^{n+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1+z)z + m/(m+1)^2}\; dz \\ = -(m+1)^2 \left(-\frac{m}{(m+1)^2}\right)^{n+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1+z)z(m+1)^2 + m}\; dz \\ = -\frac{(m+1)^2}{m-1} \left(-\frac{m}{(m+1)^2}\right)^{n+1} \\ \times \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \left(\frac{1}{1+z(m+1)} - \frac{1}{m+z(m+1)}\right)\; dz$$

Extracting coefficients from this we obtain $$-\frac{(m+1)^2}{m-1} \left(-\frac{m}{(m+1)^2}\right)^{n+1} \left((-1)^n (m+1)^n -\frac{1}{m} (-1)^n \frac{(m+1)^n}{m^n} \right).$$

This is $$-\frac{(-m)^{n+1}}{m-1} \frac{1}{(m+1)^{2n}} \left((-1)^n (m+1)^n -\frac{1}{m} (-1)^n \frac{(m+1)^n}{m^n} \right) \\ = \frac{m^{n+1}}{m-1} \frac{1}{(m+1)^{2n}} \left((m+1)^n -\frac{1}{m} \frac{(m+1)^n}{m^n} \right) \\ = \frac{m^{n+1}}{m-1} \frac{1}{(m+1)^n} \left(1 -\frac{1}{m} \frac{1}{m^n} \right).$$

We have to multiply by $(m+1)^n$ to recover the sum that we sought to evaluate, getting $$\frac{m^{n+1}}{m-1} \left(1 -\frac{1}{m} \frac{1}{m^n} \right) = \frac{m^{n+1} - m^{n+1} / m / m^n}{m-1} = \frac{m^{n+1} - 1}{m-1}.$$

Apparently this method is due to Egorychev.

Addendum. Another approach is to follow Wilf and introduce an ordinary generating function, namely $$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n {n-k\choose k} w^k.$$ This becomes $$f(z) = \sum_{k\ge 0} w^k \sum_{n\ge k} {n-k\choose k} z^n = \sum_{k\ge 0} w^k \sum_{n\ge 2k} {n-k\choose k} z^n \\= \sum_{k\ge 0} w^k \sum_{n\ge 0} {n+2k-k\choose k} z^{n+2k} = \sum_{k\ge 0} w^k z^{2k} \sum_{n\ge 0} {n+k\choose k} z^n \\= \sum_{k\ge 0} w^k z^{2k} \frac{1}{(1-z)^{k+1}} = \frac{1}{1-z} \frac{1}{1-wz^2/(1-z)} \\ = \frac{1}{1-z-wz^2}.$$ In the present case we have $$w = -\frac{m}{(m+1)^2}$$ so $f(z)$ becomes $$\frac{1}{1-z+m z^2/(m+1)^2} = \frac{(m+1)^2}{(m+1)^2 - z(m+1)^2 + m z^2}.$$ This is $$\frac{(m+1)^2}{m^2-1} \left(\frac{1}{z-m-1} - \frac{m}{mz-m-1} \right) = \frac{m+1}{m^2-1} \left(\frac{1}{z/(m+1)-1} - \frac{m}{mz/(m+1)-1} \right).$$ Extracting coefficients we obtain once more $$\frac{m+1}{m^2-1} \left(-\frac{1}{(m+1)^n} + \frac{m^{n+1}}{(m+1)^n}\right).$$ Multiply by $(m+1)^n$ to recover the original sum, getting $$\frac{1}{m-1} \left(-1 + m^{n+1}\right) = \frac{m^{n+1}-1}{m-1}.$$

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First notice that $\frac {m^{n+1}-1}{m-1}=1+m+\ldots +m^n$, which every summand is just the numbers of functions from $[i]$ to $[m]$. But that is equivalent to some functions $f$ from $[n]$ to $[m+1]$ where from some $i$, for $i<j\leq n$, $f(j)=m+1$, thats the right hand side. For the left hand side, we can use inclusion exclusion by the fact that those functions are all the functions of $[m+1]^{[n]}$ minus some functions with the property that there exists two index $i<j$ such that $f(i)=m+1$ and $f(j)\in [m]$. So suppose $P_i=\{f: f(i)=m+1 \wedge \exists j(j>i),f(j)\in[m]\}$. So you want $$|[m+1]^{[n]}\setminus \bigcup P_i|=\sum _{i=0}^n(-1)^i\sum _{a_1<a_2<\ldots < a_i}|\bigcap _{j=1}^i P_{a_j}|,$$ where $|P_i|=m*(m+1)^{n-2}C_i$, Because $i$ goes to $m+1$, we just want to see where goes the $j$ of the $P_i$, and it has $m$ possible choices and the rest of the elements (n-2) has $m+1$ choices, and the $C_i$ represents the possible choices to put the $j$ of the definition of $P_i$.
$|P_{i_1}\cap P_{i_2}|=m^2(m+1)^{n-4} C_{i_1}$, where $C_{i_1}$ represents choices to put the $j$ of the definition of $P_{i_1}$.

I think that you just need one position to put the $j$ of the first element of the chosen by the inclusion exclusion (i.e. $a_1$). So, $$|[m+1]^{[n]}\setminus \bigcup P_i|=\sum _{i=0}^n(-1)^im^i(m+1)^{n-2i}\sum _{a_1<a_2<\ldots < a_i}C_{a_1}.$$ Now, as the inclusion exclusion choose $k$ elements you have $n-k$ possibilities to put the $j$ and, if you decided to put that $j$ in position p, you have ${p-1\choose k-1}$ options to put the rest of the $a_i$'s. So by iterating all possibilities $\sum _{i=0}^{n-k-1} {i\choose k-1}={n-k \choose k}$. So you get the equality by double counting.

Hope it helps.

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  • $\begingroup$ Phica, I composed and posted my answer before seeing yours. I think we're doing something similar, but I have a hard time parsing your approach (my fault, not yours). $\endgroup$ – Barry Cipra Oct 28 '14 at 23:00
  • $\begingroup$ Barry, do not worry. It is cool to see different combinatorial approaches, even more when we are in the middle of two great calculus answers :P. I will try to see why we are doing similar things. $\endgroup$ – Phicar Oct 28 '14 at 23:16
  • $\begingroup$ Phicar, I just realized my inclusion-exclusion doesn't work. $\endgroup$ – Barry Cipra Oct 29 '14 at 0:08
  • $\begingroup$ Hi Phicar, could you please explain a little bit this particular line "I think that you just need one position to put the j of the first element of the chosen by the inclusion exclusion (i.e. a1)". $\endgroup$ – arindam mitra Oct 29 '14 at 8:38
  • $\begingroup$ $|P_{i_1} \bigcap P_{i_2}|=m^2(m+1)^{n−4}C_{i_2}$, where $C_{i_1}$ represents choices to put the j of the definition of $P_{i_1}$. did you accidentally put $C_{i_2}$ instead of $C_{i_1}$? $\endgroup$ – arindam mitra Oct 29 '14 at 8:41
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This is not a combinatorial proof but I still find it rather nice. Let us replace $m$ by $x$ to get a more general polynomial identity.

We can notice that \begin{equation*} (-1)^{k} \binom{n-k}{k} x^{k}(x+1)^{n-2k} \end{equation*} is the coefficient of $y^{k}$ in $(1+x-xy)^{n-k}$. Consequently the left hand of the identity is the coefficient of $y^{n}$ in \begin{equation*} \sum_{k=0}^{n} (y+xy-xy^{2})^{n-k}=\sum_{k=0}^{n} (y+xy-xy^{2})^{k}. \end{equation*}

For $k > n$, $(y+xy-xy^{2})^{k}$ makes no contribution to the coefficient of $y^{n}$, so we can extend the summation to infinity. Hence the left hand side equals \begin{equation*} \frac{1}{n!} \frac{\partial}{\partial^{n} y} \frac{1}{1-y-xy+xy^{2}} \end{equation*} evaluated in $y=0$. We have \begin{align} \enspace &\frac{1}{n!} \frac{\partial}{\partial^{n} y} \frac{1}{1-y-xy+xy^{2}}\\ = \enspace & \frac{1}{n!}\frac{\partial}{\partial^{n} y} \left(\frac{1}{1-y} \frac{1}{1-xy}\right) \\ = \enspace & \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k}\left(\frac{\partial}{\partial^{k} y}\frac{1}{1-y}\right)\left(\frac{\partial}{\partial^{n-k} y} \frac{1}{1-xy}\right) \end{align} But \begin{align*} &\frac{\partial}{\partial^{k} y}\frac{1}{1-y}=\frac{k!}{(1-y)^{k+1}} \\ &\frac{\partial}{\partial^{k} y}\frac{1}{1-xy}=\frac{x^{k}k!}{(1-xy)^{k+1}}. \end{align*} When $y=0$ the denominators become $1$ and the numerators simplifly with the binomial coefficient, so that indeed the left hand side equals $\displaystyle\sum_{k=0}^{n}x^{k}$.

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