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I am attempting to show that:

$$ \sum\limits_{q=0}^Q\binom{q + d - 1}{d - 1} = (Q + 1)d^Q $$

But I am totally lost. I was hoping I could use something as simple as the binomial theorem, but I quickly get stuck. Any help is greatly appreciated...

EDIT: Thanks to @stochasticboy321 for your answer! I see now that the equality cant hold... Im now trying to prove by induction a weaker, but as useful claim for my purpose, that: $$ \sum\limits_{q=0}^Q\binom{q + d - 1}{d - 1} \leq (Q + 1)d^Q $$ for $d > 0$ and $Q > 0$. But Ive gotten stuck once more :D.

Base case, $Q = 1$: $$ \binom{d - 1}{d - 1} + \binom{1 + d - 1}{d - 1} \leq 2d \implies 1 + d \leq 2d $$

Assume $\sum_{q=0}^k\binom{q + d - 1}{d - 1} \leq (k + 1)d^k$, prove:

$$ \sum\limits_{q=0}^{k+1}\binom{q + d - 1}{d - 1} \leq (k + 2)d^{k + 1} \\ \implies \sum\limits_{q=0}^k\binom{q + d - 1}{d - 1} + \binom{k + d}{d - 1} \leq (k + 2)d^{k + 1} \\ \implies (k + 1)d^k + \binom{k + d}{d - 1} \leq (k + 2)d^{k+1} $$ My strategy so far has been trying to expand $\binom{k + d}{d - 1}$ to its factorial form and reduce, but I really dont know what Im doing...

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  • $\begingroup$ This is not correct. Consider $d=2$. Then $LHS = \sum_0^Q q+1 = \frac{1}{2}(Q+1)(Q+2) \neq (Q+1)2^Q = RHS$. Indeed, wolfram alpha gives the closed form $\frac{(Q+1)}{d}\binom{d+Q}{d-1}$. $\endgroup$ – stochasticboy321 Jan 23 '17 at 0:11
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As noted in the comments, the identity asked is incorrect, and the correct identity is $$\sum\limits_{q=0}^Q\binom{q + d - 1}{d - 1} = \frac{Q+1}{d} \binom{Q+d}{d-1} = \binom{Q+d}{d}.$$ The following is a combinatorial proof of the same. Note that, by stars and bars, $\binom{q+d-1}{d-1}$ is the number of non-negative integer solutions to the equation $x_1+x_2+ \dots + x_d = q$. Thus, the summation is the number of non-negative integer solutions to $x_1+x_2+\dots + x_d \le Q$. But each non-negative integer solution of $x_1+x_2+\dots + x_d \le Q$ can be identified with a non-negative integer solution of $x_1+x_2+\dots + x_d +s = Q$, and the latter has $\binom{Q+d}{d}$ solutions. QED.


Regarding you edit - it's much easier to go inductively in $d$.

Claim: For $Q,d \ge 1, \binom{Q+d}{d} \le (Q+1)d^Q.$

Pf. Note that for every $Q$, $\binom{Q+0}{0} = Q \le Q+1,$ and thus the result holds for $d = 1$. We assume the result for $(Q,d)$. Now, for $Q,d \ge 1$,\begin{align*} (d+1)^{Q+1} \overset{(a)}\ge& d^{Q+1} + (Q+1)d^Q = d^Q(Q+d+1)\\ \iff (Q+1)(d+1)^Q \ge& \frac{(Q+1)d^Q (Q+d+1)}{d+1} \\ \overset{(b)}\implies (Q+1)(d+1)^Q \ge& \binom{Q+d}{d} \frac{Q+d+1}{d+1} = \binom{Q+d+1}{d+1} \end{align*}

where inequality $(a)$ is because each term in the binomial theorem expansion of $(d+1)^{Q+1}$ is positive, and implication $(b)$ is due to the induction hypothesis. QED.

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Algebraic Proof
Using formulas from GouldBk.pdf http://www.dsi.dsi.unifi.it/~resp/GouldBK.pdf
Partial sum ((P) in GouldBk.pdf page 14)
${\displaystyle \sum_{k=0}^{n}f_{k}=\left[t^{n}\right]\frac{f(t)}{1-t}}$
Let's rewrite the sum as
${\displaystyle \sum_{q=0}^{Q}\left(\begin{array}{c} q+d-1\\ q \end{array}\right)}$
and apply (1.49 on page 49)
${\displaystyle =\left(\begin{array}{c} d+Q\\ Q \end{array}\right)}$
Which matches the previous solution.
I couldn't do better than the 1.49 proof, which uses (BC4)
Proof (directly lifted):
${\displaystyle \sum_{q=0}^{Q}\left(\begin{array}{c} d-1+q\\ q \end{array}\right)=\left[t^{Q}\right]\frac{1}{\left(1-t\right)^{\left(d-1+1\right)}\left(1-t\right)}}=\left[t^{Q}\right]\frac{1}{\left(1-t\right)^{\left(d+1\right)}}=\left(\begin{array}{c} d+Q\\ Q \end{array}\right)$

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