How to apply:$\sum\limits^n_{k=1}f(k)=\sum\limits^n_{k=1}f(n+1-k)$ While reading a solution, I found this :$$\sum\limits^n_{k=1}f(k)=\sum\limits^n_{k=1}f(n+1-k)$$ I found it intriguing. What is the application of this? Because at the first look, it looks like it makes things more complicated. At the first look it looked like the property of integrals that $\int^b_af(x)dx=\int^b_af(a+b-x)dx$. Is there really any resemblance? **My attempt at proof:** We can write the LHS as: $$f(1)+f(2)+\dots+f(n)$$ And we can write the RHS as: $$f(n)+f(n-1)+...+f(1)$$ Thus LHS=RHS.

Imagine you stumble upon something that looks like your RHS, i.e. $\sum_{k=1}^n f(n+1-k)$, at some point during your research/homework. Then apply the formula to it and you get that it equals $\sum_{k=1}^n f(k)$. Thus you made your life easier (according to yourself, the latter expression looks less complicated than what you first stumbled upon)

PS: your proof is, of course, correct, but depending on your level it does not have the sufficient level of stringency. Why can you switch the order of sums? Why is that expression with the $\Sigma$ on the RHS equal to the expression with the dots, etc.