# proof by using induction

I'm trying to prove the following by induction but I'm stuck.

$$x_1 = 1, x_2 = 2, x_n=\frac{1}{2}(x_{n-1}+x_{n-2})$$.

Show that: $$x_n-x_{n+1} = \frac{(-1)^n}{2^{n-1}}$$

I proved the basis step, but I'm stuck in the inductive step. I tried going from L.H.S and take as $$x_{n+2}$$ as common divisor and had $$(x_{n-1} - 1)$$. I didn't know where to go from there..

• I've just edited with what I believed to be what you meant to write, please correct me if I'm wrong. Feb 25 '20 at 0:09
• @DavidPeterson When editing, please do so thoroughly. Feb 25 '20 at 0:10
• @Bernard You too. Feb 25 '20 at 0:12
• I'm sorry ,I only edit what I understand. I can see there's a problem in the last sentence, but I don't know what it should really be. Feb 25 '20 at 0:14
• @amWhy And perhaps you? Check out your subscripts in the last paragraph (which I had fixed). Feb 25 '20 at 0:14

\begin{aligned} x_{n+1} - x_{n+2} & = x_{n+1} - \frac{1}{2}(x_{n+1} + x_{n}) \\ & = x_{n+1} - \frac{x_{n+1}}{2} - \frac{x_{n}}{2} \\ & = \frac{1}{2}(x_{n+1} - x_{n}) \\ & = -\frac{1}{2}(x_{n} - x_{n+1}) \end{aligned}\tag{1}\label{eq1A}
• I tried doing that. i've substituted $x_n$ value but I'm not sure what to cancel with $x_{n+1}$ I feel so dumb because it seems so simple. thanks in advance. Feb 25 '20 at 0:34
• @raydiiii I'm not sure what you mean by "not sure what to cancel with $x_{n+1}$". Note in my ($1$) equation, if you substitute for $x_n - x_{n+1}$ what you have by the assumption in the inductive step, i.e., that $x_n - x_{n+1} = \frac{(-1)^n}{2^{n-1}}$, you'll get that $x_{n+1} - x_{n+2} = -\frac{1}{2}\left(\frac{(-1)^n}{2^{n-1}}\right) = \frac{(-1)^{n+1}}{2^{n}} = \frac{(-1)^{n + 1}}{2^{(n+1)-1}}$. Do you see this is in the same form as your assumption, but with $n$ replaced with $n + 1$? Feb 25 '20 at 0:38