Establishing formula from recurrence Can anyone tell me how do we establish a formula from a given recurrence relation?
Take the example of $f(n) = 2f(n-1) + 1$, $n \in \mathbb{Z^+}$, $f(1) = 1$
When I write out the first few values, it turns out the formula for $n$th term is $2^n-1$, but how do we get at this? I suppose that this method of observation won't work for harder ones, like $f(n) = -4f(n-1) + 3$, $n \in \mathbb{Z^+}$, $f(1) = 1$
I know methods of Lagrange interpolation and finite differences, but they won't work here, since the formula is exponential (finite difference table looks very interesting though).
 A: As you can see from Git Gud’s link, there are many techniques for finding closed forms for recurrences. Some are of quite limited applicability, while others (e.g., generating functions) are much more powerful. Here’s a technique of very limited applicability that handles recurrences of the type $f(n)=af(n-1)+b$ very nicely; I’ll illustrate it with the recurrence
$$\begin{align*}
f(n)&=-4f(n-1)+3\quad\text{for}\quad n>1\\
f(1)&=1\;.
\end{align*}\tag{1}$$
The idea is to make a substitution to replace $f$ by a function $g$ that grows exponentially. Let $g(n)=f(n)-d$ for some as yet undetermined constant $d$, so that $f(n)=g(n)+d$. Substitute into the recurrence in $(1)$ to get
$$g(n)+d=-4\big(g(n-1)+d\big)+3=-4g(n-1)+3-4d\;,$$
which simplifies to $g(n)=-4g(n-1)+3-5d$. Now choose $d$ so that $3-5d=0$, i.e., $d=\frac35$; then $g(n)=-4g(n-1)$, $g(1)=f(1)-d=\frac25$, and we can replace $(1)$ with 
$$\begin{align*}
g(n)&=-4g(n-1)\quad\text{for}\quad n>1\\
g(1)&=\frac25\;.
\end{align*}\tag{2}$$
But $(2)$ is trivial: it has the closed form $g(n)=\dfrac25(-4)^{n-1}$. It follows that
$$f(n)=g(n)+d=\frac25(-4)^{n-1}+\frac35=\frac{3-(-1)^n2^{2n-1}}5\;.$$
