Define $1+x+x^2+\cdots+x^n$ recursively

Question

Let $x \neq 1$ be a real number. Define the following sum recursively: $$ 1+x+x^2+\cdots+x^n $$

My attempt:

Using summation notation we can write the sum as $\sum_{i=0}^{n} x^i$. Now define $$ \sum_{i=0}^{0}x^i:=x^0=1 \quad \text{and} \quad \sum_{i=0}^{n}x^i:=\sum_{i=0}^{n-1}x^i+x^n $$ It seems to me correct because if we use the definition for $n=3$, then $$ \sum_{i=0}^{3}x^i=\sum_{i=0}^{2}x^i+x^3=(\sum_{i=0}^{1}x^i+x^2)+x^3=((\sum_{i=0}^{0}x^i+x)+x^2)+x^3=1+x+x^2+x^3 $$ In other words, we reach the base case. However, I'm unsure if this is correct. I might have misunderstood what a recursive definition is.

Answer 1

As I mentioned as a comment, we can define the recurrence:

\begin{equation*} f(n) = \left\{ \begin{array}{ll} x^n + f(n-1) & \quad n > 0 \\ 1 & \quad n = 0 \end{array} \right. \end{equation*}

for all $n \in \mathbb{N}_0$.

Answer 2

I think this is the best you can do: $a_n=x\cdot a_{n-1} + 1$

With $a_0=1$

Answer 3

$$f_n(x) = 1 + x + \ldots + x^n.$$

It can be rewriten as

$$f_n(x) = 1 + x\cdot\left(1 + \ldots + x^{n-1}\right),$$

or

$$f_n(x) = 1 + x\cdot f_{n-1}(x)$$

for $n>0$ and

$$f_0(x) = 1.$$