Binomially Amazing! 4

I'm using both differentiation and binomial theorem here.

$\textbf{Solution}:$

Let $f(x) := \sum^{1729}_{n=0} {1729 \choose n} x^{1729} = (1+x)^{1729}$ . Clearly, putting $x=1$ into $f(x)$ yields

\begin{aligned} \sum^{1729}_{n=0} {1729 \choose n} = 2^{1729}\tag{0} \end{aligned}

Differentiating $f$ w.r.t. $x$ once and twice respectively and putting $x=1$ yield

\begin{aligned} \sum^{1729}_{r=0} r {1729 \choose n} &= 1729 \cdot 2^{1728}\tag{1} \\ \sum^{1729}_{r=0} r(r-1) {1729 \choose n} &= 1729 \cdot 1728 \cdot 2^{1727}\quad \quad \tag{2}\\ \end{aligned}

We can expand the left-hand side of (2): $\begin{aligned} \sum^{1729}_{r=0} r^2 {1729 \choose n} - \sum^{1729}_{r=0} r {1729 \choose n} &= 1729 \cdot 1728 \cdot 2^{1727}\\ \sum^{1729}_{r=0} r^2 {1729 \choose n} &= \sum^{1729}_{r=0} r {1729 \choose n} + 1729 \cdot 1728 \cdot 2^{1727}\\ &= 1729 \cdot 2^{1728} + 1729 \cdot 1728 \cdot 2^{1727}\quad \quad \quad \quad \text{(1) is used} \end{aligned}$

Now the LHS of the given equation

$\begin{aligned}\sum^{1729}_{n=0} \left(n - \frac{1729}{2}\right)^2 {1729 \choose n} \left(\frac{1}{2}\right)^{1729}\end{aligned}$

can be broken down into

$\begin{aligned}\left(\frac{1}{2}\right)^{1729} \sum^{1729}_{n=0} n^2 {1729 \choose n} - 1729 \left(\frac{1}{2}\right)^{1729} \sum^{1729}_{n=0} n {1729 \choose n} + \left(\frac{1729}{2}\right)^2 \left(\frac{1}{2}\right)^{1729}\sum^{1729}_{n=0} {1729 \choose n}\end{aligned}$

Putting the above results into this equation yields $\begin{aligned} &\left(\frac{1}{2}\right)^{1729} \left(1729 \cdot 2^{1728} + 1729 \cdot 1728 \cdot 2^{1727} \right)- 1729 \left(\frac{1}{2}\right)^{1729} \cdot 1729 \cdot 2^{1728}+ \left(\frac{1729}{2}\right)^2 \left(\frac{1}{2}\right)^{1729}\cdot 2^{1729}\\ =& \frac{1729}{2} + \frac{1729 \times 1728}{4} - \frac{1729^2}{2} + \frac{1729^2}{4}\\ =&\frac{1729}{4}\end{aligned}$

and hence $A + B = 1733$ where we have assumed $A$ and $B$ are positive coprime integers. $\Box$

I will use Binomial Theorem , in particular $\sum_{n=0}^{m}{m \choose n}(\frac{1}{2})^{n}(\frac{1}{2})^{m-n}=(\frac{1}{2}+\frac{1}{2})^m=1 \\$

And the property $k{n \choose k}=n{n-1 \choose k-1}$

Let

$S=\sum_{n=0}^{1729}(n^2-1729n+\frac{1729^2}{4}){1729 \choose n}(\frac{1}{2})^{1729}$

We can split $S$ in three sums:

$S_1=\sum_{n=0}^{1729}n^2{1729 \choose n}(\frac{1}{2})^{1729} \\ S_2=\sum_{n=0}^{1729}n{1729 \choose n}(\frac{1}{2})^{1729} \\ S_3=\sum_{n=0}^{1729}{1729 \choose n}(\frac{1}{2})^{1729} \\ S=S_1-1729S_2+\frac{1729^2}{4}S_3$

$\\ \textbf{Manipulating } S_3 \\ \sum_{n=0}^{1729}{1729 \choose n}(\frac{1}{2})^{n}(\frac{1}{2})^{1729-n}=1$

$\\ \textbf{Manipulating } S_2 \\ \sum_{n=0}^{1729}n{1729 \choose n}(\frac{1}{2})^{1729}\\ =\frac{1729}{2}\sum_{n=0}^{1729}{1728 \choose {n-1}}(\frac{1}{2})^{1728}\\ =\frac{1729}{2}\sum_{n=-1}^{1728}{1728 \choose {n}}(\frac{1}{2})^{1728}\\ =\frac{1729}{2}\sum_{n=0}^{1728}{1728 \choose {n}}(\frac{1}{2})^{n}(\frac{1}{2})^{1728-n}=\frac{1729}{2}(1)=\frac{1729}{2}$

$\\ \textbf{Manipulating } S_1 \\ \sum_{n=0}^{1729}n^2{1729 \choose n}(\frac{1}{2})^{1729}\\ =1729\sum_{n=0}^{1729}(n-1+1){1728 \choose {n-1}}(\frac{1}{2})^{1729}\\ =1729\Big[\sum_{n=0}^{1729}(n-1){1728 \choose {n-1}}(\frac{1}{2})^{1729}+\sum_{n=0}^{1729}{1728 \choose {n-1}}(\frac{1}{2})^{1729}\Big]\\ =1729\Big[1728\sum_{n=0}^{1729}{1727 \choose {n-2}}(\frac{1}{2})^{1729}+\sum_{n=-1}^{1728}{1728 \choose {n}}(\frac{1}{2})^{1729}\Big]\\ =1729\Big[1728\sum_{n=-2}^{1727}{1727 \choose n}(\frac{1}{2})^{1729}+\frac{1}{2}\sum_{n=0}^{1728}{1728 \choose {n}}(\frac{1}{2})^{n}(\frac{1}{2})^{1728-n}\Big]\\ =1729\Big[\frac{1728}{4}\sum_{n=0}^{1727}{1727 \choose n}(\frac{1}{2})^{n}(\frac{1}{2})^{1727-n}+\frac{1}{2}\Big]\\ =1729\Big[\frac{1728}{4}+\frac{1}{2}\Big]=\frac{1730 \times 1729}{4}$

$\\ \textbf{Now, for } S \\ S=\frac{1730 \times 1729}{4}-\frac{1729 \times 1729}{2}+\frac{1729^2}{4}=\boxed{\frac{1729}{4}} \\ \text{So, the answer is \$1729+4=\boxed{1733}\$}$