# $f(n) = k^n$ for some $n$?

To be specific, there have been questions on this topic:

n = 0 to 5 (inclusive)

n = 0 to 6 (inclusive)

This wiki contains spoliers about these questions, so you are recommended to do them first before reading the rest of this note.

In these questions: Actually, please do those questions first.

In these questions, $k=2$ and $n = 0$ to $n=5$ and $n=0$ to $n=6$.

There's a lot of similarities, and even the answer is amazing, but I shall not spoil it.

Both questions are in the form "$f(n)$ is a $p$-degree polynomial. When $q = 0,1,2,\ldots,p$, $f(q) = k^q$. What is $f(x)$?"

These questions are a special case of this form, with $k = 2$ and x = 2*q+1. The answer is also very similar, f(x) = 2^(2*q).

I would like to know if this can be generalised for q and also, k, and if so, what values of x can create these values.

I'll work on it when I have time. Which doesn't come a lot.

Note by Aloysius Ng
3 years, 7 months ago

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It can be done.

${ 3 }^{ n }={ 2 }^{ 0 }\left( \begin{matrix} n \\ 0 \end{matrix} \right) +{ 2 }^{ 1 }\left( \begin{matrix} n \\ 1 \end{matrix} \right) +{ 2 }^{ 2 }\left( \begin{matrix} n \\ 2 \end{matrix} \right) +\cdots +{ 2 }^{ n-1 }\left( \begin{matrix} n \\ n-1 \end{matrix} \right) +2^{ n }\left( \begin{matrix} n \\ n \end{matrix} \right)$

- 3 years, 6 months ago