Telescoping is a method for finding the sum of certain series.
If we have a sequence \(a_0\), \(a_1\), \(a_2\), …, and we can find a function \(f(n)\) such that \(a_n=f(n+1)-f(n)\) for every \(n\), then \[\sum_{i=0}^n a_i = \sum_{i=0}^n f(i+1)-f(i) = f(n+1)-f(0)\] (because all the other terms cancel).
But finding such a function explicitly may be difficult.
As an example, for the sequence \(a_n=n\), we note that \((n+1)^2-n^2=2n+1\), and \((n+1)-n=1\), so if we take \(f(n)=\frac{1}{2}(n^2-n)\), we will have \[\begin{align*} f(n+1)-f(n) &=\tfrac{1}{2}\bigl((n+1)^2-(n+1)\bigr) - \tfrac{1}{2}(n^2-n)\\ &=\tfrac{1}{2}(n^2+n) - \tfrac{1}{2}(n^2-n)\\ &=n\\ &=a_n \end{align*}\] so that \[\begin{align*} \sum_{i=0}^n i&=f(n+1)-f(0)\\ &=\tfrac{1}{2}\bigl((n+1)^2-(n+1)\bigr)-\tfrac{1}{2}(0^2-0)\\ &=\tfrac{1}{2}n(n+1). \end{align*}\]
