Mimicking amWhy's proof yields a powerful result
on telescoping sums:
$\bbox[1px,border:1px solid #c00]{\bbox[5px,border:1px solid #c00]{\displaystyle\quad\sum_{i=1}^n f(i) = g(n)\iff \color{#c00}{g(1) = f(1)}\,\ {\rm and}\,\ \color{#0a0}{g(n\!+\!1)-g(n) = f(n\!+\!1)}\,\ {\rm for\ all}\ n \ge 1}}$
Base case:
Let $n=1$ and test: $$\sum_{i=1}^1 f(i) = f(1) \color{#c00}{=?}\ g(1)$$
The $ $ Base Case $\ \,n = 1\ $ holds true $\iff \color{#C00}{f(1) = g(1)}$
Induction Hypothesis:
- Assume that it is true for $\, n = k$: assume that $$\sum_{i=1}^k f(i) = g(k).$$
Inductive Step:
Prove, using the Induction Hypothesis as a premise, that $$\sum_{i=1}^{k+1}f(i)=\left(\sum_{i=1}^k f(i)\right) + f(k\!+\!1) = g(k) + f(k\!+\!1) \color{#0a0}{=?}\ g(k\!+\!1).$$
The Inductive Step from $k$ to $k+1$ is true $\iff \color{#0a0}{ g(k\!+\!1) - g(k) = f(k\!+\!1)}$
This theorem reduces the inductive proof to simply verifying the $\rm\color{#c00}{RHS}$ $\rm\color{#0a0}{equalities}$, which is trivial polynomial arithmetic when $f(n),g(n)$ are polynomials in $n,\,$ so trivial it can be done purely mechanically by a high-school student (or computer). No insight (magic) is required - no rabbits need be pulled from a hat.
The above theorem is an example of telescopy, also known as the Fundamental Theorem of Difference Calculus, depending on context. You can find many more examples of telescopy and related results in other answers here.