Generalize a sum given first elements Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as
$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$
generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)
$$\sum_{n=0}^{\infty_+}(1+r)^{-n}f(n)$$
I think I could simplify the summator but I can't get my head around generalizing $f(n)$. What should be the approach to this? Thanks.
 A: One possible generalized $f$  is $$f(n)=\sum_{i=0}^{n} (i+1){\rho}^{n-i}$$.
There can be many $f$. But here it fits.
A: $f$ can be represented as a reccurence relation by defining $f(0)=1$.
$$f(r)=\rho f(r-1)+(r+1)\forall r\in\mathbb{Z^{+}}$$
A: Try $f(n)=(n+1)+(n)\rho+(n-1)\rho^2+\cdot\cdot\cdot+(1)\cdot\rho^n=\sum\limits_{i=1}^{n+1} i\rho^{n+1-i} = \sum\limits_{i=0}^{n} (i+1)\rho^{n-i}$
$f(n)$ has $n+1$ terms in its sum where the sum of the coefficient of each term and the power of $\rho$ is $n+1$.
A: I'm not sure what you're trying to do with your example, but one way of expressing $f(n)$ with a general formula is $$f(n) = \sum_{i=0}^{n} (n-i+1)\rho^i.$$
A: I would like to add one more approach.
Consider
$$
f(n) = (n+1) + n \rho + (n-1) \rho^2 + \cdots + 1 \rho^n = \sum_{k=1}^{n+1} k \rho^{n+1-k}.
$$
Obviously $f(0)= 1$, $f(1) = 2 + \rho$ and so on.
Now make this function a bit more complicated
$$
f(n,x) = (n+1) x^n + n x^{n-1} \rho + (n-1) x^{n-2} \rho^2 + \cdots + 1 \rho^n.
$$
You may see that $f(n,1) = f(n)$.
Let's represent our new function as a derivative and do few transformations
$$
\begin{align}
f(n,x) 
&= \frac{\partial}{\partial x} \left(x^{n+1} + x^{n} \rho + x^{n-1} \rho^2 + \cdots + x \rho^n\right) =\\
&= \frac{\partial}{\partial x} \left( x \left[x^{n} + x^{n-1} \rho + x^{n-2} \rho^2 + \cdots + \rho^n \right] \right) =\\
&= \frac{\partial}{\partial x} \frac{x (x - \rho) (x^{n} + x^{n-1} \rho + x^{n-2} \rho^2 + \cdots + \rho^n)}{x - \rho} =\\
&= \frac{\partial}{\partial x} \frac{x (x^{n+1} - \rho^{n+1})}{x - \rho}
\end{align}
$$
Doing all the boring stuff with derivatives and setting $x=1$, at the end of the day we get
$$
f(n) = \frac{(n+1) - \rho(n+2) + \rho^{n+2}}{(\rho - 1)^2}
$$
One may check that
$$
f(0) = \frac{1 - 2\rho + \rho^2}{(\rho - 1)^2} = 1,
$$
$$
\begin{align}
f(1) &= \frac{2 - 3\rho + \rho^3}{(\rho - 1)^2}
      = \frac{2 - 4\rho + 2\rho^2 + \rho^3 - 2\rho^2 + \rho}{(\rho - 1)^2} =\\
     &= \frac{2(1-\rho)^2+ \rho(\rho - 1)^2}{(\rho - 1)^2}
      = 2 + \rho
\end{align}
$$
and so on.
A: It is easy to define $f$ recursively. This is easier to see when we reverse the order of the terms:
\begin{align*}
f(0) &= 1 \\
f(1) &= \rho + 2 = \rho f(0) + 2 \\
f(2) &= \rho^2 + 2\rho + 3 = \rho f(1) + 3 \\
f(3) &= \rho^3 + 2\rho^2 + 3\rho + 4 = \rho f(2) + 4
\end{align*}
We can define $f$ as $f(0) = 1$, $f(n) = \rho f(n-1) + (n + 1)$ for $n \geq 1$.
Paras Khosla’s answer shows a different way of writing this definition.
