0

Suppose $~a_0 = 1~$ and $~a_n = 2a_{n−1} + 3~$ for all $~n ≥ 1$. Using proof by induction, prove that the formula for an for all $~n ≥ 0~$ is given by $~a_n = 2^{n+2} − 3$.

How can I solve this question?

Calvin Lin
  • 68,864
  • Hint: Show that the case for $n=0$ holds. Then, assuming the formula holds for $n=k$, show that the formula for $n=k+1$ holds. This will likely require you to use the actual recurrence to express $a_{k+1}$ in terms of $a_k$. If you want some more detailed help, please [edit] your post to include some details on your own attempts and understanding of this problem. – PrincessEev Nov 21 '19 at 06:42

1 Answers1

0

Hint: $a _ n + 3 = 2 ( a_{n-1} +3) $.

Calvin Lin
  • 68,864