Maximum difference between 2 consecutive terms in a row of Pascal's triangle

Question

For a given $n$ and $n > 1$, when does the below attain a maximum? $$F(i) = {n \choose i+1} - {n \choose i} $$

How I approached it? I started by considering mapping $g(i)$ from $i$ to ${n \choose i}$ for a given $n$. I observed that $F(i)$ describes the change in $g(i)$ when $i$ is changed by $1$. Therefore, I thought there must be another $h(i)$ which describes changes in $F(i)$ on a unit change in $i$. Hence, $$h(i) = F(i+1) - F(i)$$

So I proceeded and substituted $F(i+1)$ and $F(i)$ into above function, and set $h(i) = 0$. But I found that it did not give me the right answer. I was aware that I am not using Calculus to differentiate $F(i)$, because I do not know if how to differentiate F(i).

Answer 1

$$ \begin{align} F(i)-F(i-1) &=\binom{n}{i+1}-2\binom{n}{i}+\binom{n}{i-1}\\ &=\frac{n!}{(i+1)!(n-i-1)!}-\frac{2n!}{i!(n-i)!}+\frac{n!}{(i-1)!(n-i+1)!}\\[3pt] &=\frac{n![(n-i)(n-i+1)-2(i+1)(n-i+1)+i(i+1)]}{(i+1)!(n-i+1)!}\\ &=\frac{n!\left[4i^2-4ni+(n-2)(n+1)\right]}{(i+1)!(n-i+1)!}\tag{1} \end{align} $$ The quadratic formula says that $F(i)=F(i-1)$ when $i=\frac{n\pm\sqrt{n+2}}2$. Since the value in $(1)$ is positive outside $\left[\frac{n-\sqrt{n+2}}2,\frac{n+\sqrt{n+2}}2\right]$ and negative inside, the maximum would be at $\frac{n-\sqrt{n+2}}2$. Thus, the maximum should be at the integer closest to $$ \bbox[5px,border:2px solid #C0A000]{\frac{n-1-\sqrt{n+2}}2}\tag{2} $$

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Example: $n=100$: $\frac{99-\sqrt{102}}2=44.45$
$F(45)=\binom{100}{46}-\binom{100}{45}=12022526976678817747184463840$
$\color{#C00}{F(44)=\binom{100}{45}-\binom{100}{44}=12070235417062463849355830760}$
$F(43)=\binom{100}{44}-\binom{100}{43}=11261702901086987802030559800$

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Example: $n=101$: $\frac{100-\sqrt{103}}2=44.93$
$F(44)=\binom{101}{45}-\binom{101}{44}=23331938318149451651386390560$
$\color{#C00}{F(45)=\binom{100}{45}-\binom{100}{45}=24092762393741281596540294600}$
$F(46)=\binom{100}{44}-\binom{100}{43}=22965016068927859904787344640$

Answer 2

What is $F(i+1)-F(i)$? It is $$\begin{align} \binom{n}{i+2}-2\binom{n}{i+1}+\binom{n}{i}&= \binom{n}{i}\left[\frac{(n-i)(n-i-1)}{(i+2)(i+1)}-2\frac{n-i}{i+1}+1\right] \\ &=\binom{n}{i}\frac{(n-i)(n-i-1)-2(n-i)(i+2)+(i+2)(i+1)}{(i+2)(i+1)}\\ &=\binom{n}{i}\frac{n^2-(4i+3)n+4i^2+8i+2}{(i+2)(i+1)}. \end{align} $$ This doesn't seem to factor nicely, alas. But it will change sign twice, once where $F(i)$ starts to decrease and the other time when it starts to increase (but by then $F(i)$ will be negative).

Answer 3

$$F(i) = {n \choose i+1} - {n \choose i} $$ $$F(i) = n!\left[\frac{1}{(i+1)!(n-(i+1))!}-\frac{1}{i!(n-i)!}\right] $$ $$F(i) = \frac{n!}{i!(n-(i+1))!}\left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right] $$$$F(i)=\frac{n!}{\Gamma(i+1)\Gamma(n-(i+1)+1)}\left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right] $$ So, $$F(i)=\left[\frac{n!}{\Gamma(i+1)\Gamma(n-i)}\right]\left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right] \hspace{10mm} (1)$$ Let $$G(i)=\left[\frac{n!}{\Gamma(i+1)\Gamma(n-i)}\right]; J(i) = \left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right]$$

Notes:

• $n!=\Gamma(n+1)$, i.e. we use the Gamma function.
• We want the derivative of the factorial at integers. Refer to: Derivative of a factorial
• $H_n$ is the $n^\text{th}$ Harmonic Number
• $\gamma$ is the Euler-Mascheroni constant

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On differentiating equation (1), $$F'(i)=G'(i)\cdot J(i) + G(i)\cdot J'(i)$$ You already know $G(i)$ and $J(i)$ (mentioned above), so let us find their derivatives. $$J'(i)=\frac{-1}{(i+1)^2}+\frac{-1}{(n-i)^2}$$ And $$G'(i)=\left[\frac{n!}{\Gamma(i+1)\Gamma(n-i)}\right]' =-n!\frac{\color{red}{\Gamma'(i+1)}\Gamma(n-i)+\Gamma(i+1)\color{red}{\Gamma'(n-i)}}{(\Gamma(i+1)\Gamma(n-i))^2}$$ We get the above using quotient rule for differentiation.

Since $$\Gamma'(i+1)=i!(H_i-\gamma)$$ and $$\Gamma'(n-i)=-[n-(i+1)]!(H_{n-(i+1)}-\gamma)$$ and $$\Gamma(i+1)=i!$$ and $$ \Gamma(n-i)=(n-(i+1))!$$We get: $$G'(i) =-n!\left[\frac{\left[i!(H_i-\gamma)\right](n-(i+1))!-i!\left[(n-(i+1))!(H_{n-(i+1)}-\gamma)\right]}{(\Gamma(i+1)\Gamma(n-i))^2}\right]$$ $$G'(i) =(n! \cdot i! \cdot (n-(i+1))!)\left[\frac{\left[(H_{n-(i+1)}-\gamma)\right]-\left[(H_i-\gamma)\right]}{(\Gamma(i+1)\Gamma(n-i))^2}\right]$$ That is, $$G'(i)=\left[H_{n-(i+1)}-H_i\right]\left[\frac{(n! \cdot i! \cdot (n-(i+1))!)}{(i! \cdot (n-(i+1))!)^2}\right]$$ Therefore: $$G'(i)=\left[H_{n-(i+1)}-H_i\right]\left[{n \choose i+1}\cdot(i+1)\right]$$

Finally, we get: $$\color{purple}{F'(i)=\left[\left[H_{n-(i+1)}-H_i\right]\left[{n \choose i+1}\cdot(i+1)\right]\right]\cdot\left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right]+\left[\frac{n!}{\Gamma(i+1)\Gamma(n-i)}\right]\cdot\left[\frac{-1}{(i+1)^2}+\frac{-1}{(n-i)^2}\right]}$$

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You wish to maximise $F(i)$.

Plug in the formula from the above section for $F'(i)$ set $F'(i)=0$ to get the function's critical points which you can determine to be local minima or local maxima or merely inflexion points using the second derivative test.

Hope the answer is a big enough hint for you to arrive at the answer.