Combinations: summation of combinations equalities

Question

Suppose we have two quantities

$$ A = \sum^n_{i=0}C^n_i (X_{n-i}X_{i+1} + X_iX_{n-i+1})\\ B = \sum^{n+1}_{i=0}C^{n+1}_i (X_iX_{n-i+1}), $$

where $C^n_i$ is the combination notation, and $X$ are just some numbers. How to prove $A=B$?

Answer 1

Here, I will use the following facts: $$C^n_{n+1}=C^n_{-1}=0\tag{1}$$ $$C^{n+1}_i=C^n_{i-1}+C^n_i\tag{2}$$

I will also be assuming some basic knowledge of re-indexing sums, and assuming that you meant to say $$B = \sum^{n+1}_{i=0}C^{n+1}_i (X_iX_{n-i+1}),$$ rather than $$B = \sum^{n+1}_iC^{n+1}_i (X_iX_{n-i+1}).$$

With that in mind, we see that

$$\begin{eqnarray}A &=& \sum^n_{i=0}C^n_i (X_{n-i}X_{i+1} + X_iX_{n-i+1})\\ &=& \sum^n_{i=0}C^n_iX_{n-i}X_{i+1} + \sum^n_{i=0}C^n_iX_iX_{n-i+1}\\ &=& \sum^n_{i=0}C^n_iX_{i+1}X_{n-i} + \sum^n_{i=0}C^n_iX_iX_{n-i+1}\\ &=& \sum^{n+1}_{i=1}C^n_{i-1}X_{(i-1)+1}X_{n-(i-1)} + \sum^n_{i=0}C^n_iX_iX_{n-i+1}\\ &=& \sum^{n+1}_{i=1}C^n_{i-1}X_{i}X_{n-i+1} + \sum^n_{i=0}C^n_iX_iX_{n-i+1}\\ &\overset{(1)}{=}& \sum^{n+1}_{i=0}C^n_{i-1}X_{i}X_{n-i+1} + \sum^{n+1}_{i=0}C^n_iX_iX_{n-i+1}\\ &=& \sum^{n+1}_{i=0}\left(C^n_{i-1}+C^n_i\right)X_iX_{n-i+1}\\ &\overset{(2)}{=}& \sum^{n+1}_{i=0}C^{n+1}_iX_iX_{n-i+1}\\ &=& B\end{eqnarray}$$

Please let me know if there are any steps you aren't sure of, or if you have any other questions about my answer.