f(n)=knf(n) = k^n for some nn?

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=2k=2 and n=0n = 0 to n=5n=5 and n=0n=0 to n=6n=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)f(n) is a pp-degree polynomial. When q=0,1,2,…,pq = 0,1,2,\ldots,p, f(q)=kq f(q) = k^q. What is f(x)f(x)?"

These questions are a special case of this form, with k=2k = 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.

#Algebra

Note by Aloysius Ng
5 years, 3 months ago

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Comments

It can be done.

3n=20(n0)+21(n1)+22(n2)+⋯+2n−1(nn−1)+2n(nn){ 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)

Joel Yip - 5 years, 2 months ago
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