Question

\[\text{if } \alpha+\frac{1}{\alpha}=k\\ \text{then find}\\ \alpha^n+\frac{1}{\alpha^n}\]

Answer 1

What does \[\alpha + \frac{ 1 }{ \alpha }\]equal? Do we know that? Or are we just rewriting in a more usable way?

Answer 2

i edited it,its okay now

Answer 3

a more easier question would be
\[x+\frac{1}{x}=3\\x^2+\frac{1}{x^2}=?\]

Answer 4

Right. So what we would do in that case is square \[(x + \frac{ 1 }{ x })^2 = x^2 + 2 + \frac{ 1 }{ x^2 }\]So we have\[x^2 + \frac{ 1 }{ x^2 } + 2 = 81\]So \[x^2 + \frac{ 1 }{ x^2 } = 79\]Makes sense?

Answer 5

yes okay so can we generalise that
\[x^2+\frac{1}{x^2}=3^2-2\]
and hence
\[x^n+\frac{1}{x^n}=k^n-n \]

Answer 6

Not quite. That isn't always right. Try multiplying out (x + 1/x)^3 and seeing if you get what you expect.

Answer 7

Are you sure that the problem says that you need to find\[x^n + \frac{1}{x^n}\]With the powers being n? Because that's a really hard generalization to spot - I'm not even sure I can spot it without a great deal of algebra.

Answer 8

\[(x+\frac{1}{x})^n=\sum _{k=0}^n\left(\begin{matrix}n\\ \\ k\end{matrix}\right)x^{n-k}(\frac{1}{x^{k}})=\sum _{k=0}^n\left(\begin{matrix}n\\ \\ k\end{matrix}\right)x^{n-2k}\]
\[(x+\frac{1}{x})^n-\sum _{k=1}^{n-1}\left(\begin{matrix}n\\ \\ k\end{matrix}\right)x^{n-2k}=x^n+\frac{1}{x^n}\]

Answer 9

\[\alpha^n+\frac{1}{\alpha^n}=k^n-\left(\begin{matrix}n \\ 1\end{matrix}\right)x^{n-2}+\left(\begin{matrix}n \\ 2\end{matrix}\right)x^{n-4}+....+\left(\begin{matrix}n \\ n-1\end{matrix}\right)x^{n-2(n-1)}\]