A few questions about a question involving mathematical induction 
Suppose we have a sequence of numbers defined recursively as follows:
  $$ d_n = \begin{cases} 0 & \text{if } n=1; \\ 2d_{n-1}+2 & \text{if }
n \geqslant 2. \end{cases} $$ Using mathematical induction, prove that for
  every integer $n\geqslant 1, d_n = 2^n -2$.

I have a few questions about this assignment.


*

*We need to use mathematical induction, but do we still need a basis like $n=1$, because it already says that $d_n = 0$

*Since $d_n =0$ for $n=1$ and $d_{n-1} +2$ for $n \ge 2$. And the fact that $d_n = 2^n -2$, and for the induction part $n$ needs to be $k$, do we then do this:
$$
     2^k -2 = 2(2^k-1 +2)
$$ 
or do we do 
$$
2^k -2 = 2(2^k +2)
$$

*Same question for $n = k+1$: 
$$
     2^k+1 -2 = 2(2^k +2)
$$ 
or do we do 
$$
2^k+1 -2 = 2(2^k+1 +2)
$$
Thanks in advance. Sorry if i wasnt to clear in my title. 

I don't really understand your reasoning. You know that $d_1=0$, with checks out with $d_1=2^1-2$. Now assume that the relation holds up to $n$, this is your induction hypothesis. Then $$ d_{n+1}=2d_n+2$$ And you just have to use your induction hypothesis. I'll let you go on from here.
