IRL_ABooklet_2023

If $p,q$ are polynomials of $\mathrm{deg}\le n$ with coefficients $a_k,b_k$ respectively, then the coefficients of the product $pq$ may be found in terms of those of $p$ and $q$ as follows: $$\begin{array}{rl}p(x)q(x)& =\sum _{i=0}^n{a}_i{x}^i\sum _{j=0}^n{b}_j{x}^j=\sum _{i=0}^n{a}_i{b}_j{x}^{i+j}\\ & =\sum _{k=0}^{2n}\left(\sum _{i+j=k}{a}_i{b}_j\right)x^k=\sum _{k=0}^{2n}\left(\sum _{i=0}^k{a}_i{b}_{k-i}\right)x^k\end{array}$$

To obtain expression $B$, apply this formula with $$p(x)=(1+x{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k\text{and}q(x)=(1-x{)}^n=\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})x^k.$$ Then $$\sum _{k=0}^{2n}\left(\sum _{i=0}^k(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{n}{k-i})\right)x^k=p(x)q(x)=(1-x^2{)}^n=\sum _{r=0}^n(-1{)}^r\left(\genfrac{}{}{0ex}{}{n}{r}\right)x^{2r},$$ so that, comparing coefficients, we see that, if $0\le k\le 2n$, $$\sum _{i=0}^k(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{n}{k-i})=\{\begin{array}{ll}0,& \text{if }k\text{ is odd,}\\ (-1{)}^{\frac{k}{2}}(\genfrac{}{}{0ex}{}{n}{\frac{k}{2}}),& \text{if }k\text{ is even.}\end{array}$$ In particular, taking $k=n$, $$B=\sum _{k=0}^n(-1{)}^k{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2=\{\begin{array}{ll}0,& \text{if }n\text{ is odd,}\\ (-1{)}^{\frac{n}{2}}(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}}),& \text{if }n\text{ is even.}\end{array}$$