An algebra problem by Budi Utomo

The general form of the sum (putting $n=2016$ ) is as follows:

$\begin{aligned} S & = \sum_{k=0}^{n-1} {n \choose k}{n \choose k+1} \\ & = \sum_{k=0}^{n-1} {n \choose k}{n \choose n - k -1} & \small \color{#3D99F6} \text{Vandermonde's identity: } {m+n \choose r} = \sum_{k=0}^r {m \choose k}{n \choose r-k} \\ & = {m+n \choose r} & \small \color{#3D99F6} \implies m = n, \ r = n-1 \\ & = {2n \choose n-1} \end{aligned}$

$\implies a = 2n$ and $b=n-1$ , the other $b=n+1$ which is larger.

Therefore, $a=4032$ , $b= 2015$ and $a+b=\boxed{6047}$ .

---

Reference: Vandermonde's identity

the sum is equivalent to selecting 2015 objects out of 4032 distict objects

The given expression is equal to the coefficient of $x$ or $\dfrac{1}{x}$ in the expansion of

$(1+x)^{2016} \times(1+\dfrac{1}{x})^{2016}$

Now the expression, $(1+x)^{2016} \times(1+\dfrac{1}{x})^{2016}=(1+x)^{2016} \times\dfrac{(1+x)^{2016}}{x^{2016}}=\dfrac{(1+x)^{4032}}{x^{2016}}$

We need to find the coefficient of $x$ or $\dfrac{1}{x}$ in the above expression

thus it simplifies to finding the coefficient of $x^{2017}$ or $x^{2015}$ in the numerator of the expression

this is equal to $\large{4032 \choose2017}$ and $\large{4032 \choose 2015}$ respectively

since we need to minimize $a+b$ we choose the latter and get $a+b=\boxed{6047}$

There are 2 answers 4032C2015 and 4032C2017. So a+b can be 6047 and 6049.

(C0Cr)+(C1Cr+1)+.....+(Cn-rCn)=2nCn-r=4032C2015.So a+b=6047.