An identity between summations involving a binomial expansion.

Question

Could someone give me some hint to prove that this equality is an identity?

It's not about homowork; it has arisen in a development and what I want is to go from RHS to the LHS, since the LHS is the most convenient way for the purposes for which I have developed it.

I have verified that it holds for some particular cases and I am sure that it is true in general, but I want to justify it formally. I have not deduced the LHS algebraically, but by observing particular cases.

<<<<<<<<<<<<<<<<<<<<<<<<<<< EDIT 1 >>>>>>>>>>>>>>>>>>>>>>>>>

Below I show the check for m = 4.

It can be seen that the same results are obtained (identical polynomials are generated).

What varies is the way the coefficients of each power of x are distributed. In the LHS case, these are distributed by columns, while in the RHS case they are distributed diagonally.

With the change m - n = j (suggested by Jean Marie) it is possible to match the subscripts of the deltas and the form of the binomial coefficients (combinatorial numbers), but new discrepancies are introduced (the exponents of -1 don't match and the powers of x are restricted to even exponents).

Answer 1

We show the following identity is valid \begin{align*} \color{blue}{\sum_{n=0}^m}&\color{blue}{\sum_{j=0}^n(-1)^{n+j}\binom{m-j}{n-j}\Delta_jx^n =\sum_{n=0}^m\sum_{j=0}^n(-1)^j\binom{n}{j}\Delta_{m-n}x^{m+j-n}}\tag{1} \end{align*}

We start with the left-hand side of (1). We obtain \begin{align*} \color{blue}{\sum_{n=0}^m}&\color{blue}{\sum_{j=0}^n(-1)^{n+j}\binom{m-j}{n-j}\Delta_jx^n}\\ &=\sum_{j=0}^m\sum_{n=j}^{m}(-1)^{n+j}\binom{m-j}{n-j}\Delta_jx^n\tag{2}\\ &=\sum_{j=0}^m\sum_{n=0}^{m-j}(-1)^n\binom{m-j}{n}\Delta_jx^{n+j}\tag{3}\\ &=\sum_{j=0}^m\sum_{n=0}^j(-1)^n\binom{j}{n}\Delta_{m-j}x^{m+n-j}\tag{4}\\ &\,\,\color{blue}{=\sum_{n=0}^m\sum_{j=0}^n(-1)^j\binom{n}{j}\Delta_{m-n}x^{m+j-n}}\tag{5} \end{align*} and the claim (1) follows.

Comment:

• In (2) we exchange inner and outer sum.

• In (3) we shift the index of the inner sum to start with $j=0$.

• In (4) we change the order of summation of the outer sum $j\to m-j$.

• In (5) we exchange variables $n\longleftrightarrow j$.