The question says to allow $x_1=3$ and given $x_n$:
$$x_{n+1} = \frac{1}{4}x_n + 9.$$ Use mathematical induction to prove that for all $n \in\mathbb N$: (a) $x_{n+1}> x_n$, (b) $x_n < 12$. I am not completely sure I know what's being asked here. Could someone clarify and give a hint.. I think I at least got the base case.. Compare the equation with 3 and then the equation with 4. Or is that wrong, too?
2026-03-28 07:35:10.1774683310
Confused by notation for Mathematical Induction question
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I would start first by proving part b. If $x_{n-1}<12$ then $\frac{1}{4}x_{n-1}<3$, therefore $x_n<3+9=12$.
For part a, just compute $x_{n+1}-x_n$. You will get $9-\frac{3}{4}x_n$. But now you know $x_n<12$, so $x_{n+1}-x_n>9-\frac{3}{4}12$