Maximum difference between 2 consecutive terms in a row of Pascal's triangle 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).
 A: 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).
A: $$F(i) = {n \choose i+1} - {n \choose i} $$
$$F(i) = n!\left[\frac{1}{(i+1)!(n-(i+1))!}-\frac{1}{i!(n-i)!}\right] $$
$$F(i) = \frac{n!}{i!(n-(i+1))!}\left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right]
$$$$F(i)=\frac{n!}{\Gamma(i+1)\Gamma(n-(i+1)+1)}\left[\frac{1}{(i+1)}-\frac{1}{(n-i)}\right] $$
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

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

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.
