k=n-1 in recurrence relations When using the iterative technique when solving for recurrence relations, why do we set k= n-1? What allows us to set it to this value and why do we do this?
 A: Recursion relations are sensitive to initial conditions, which is to say the state that you start in can change the outcome you receive. For example the Fibonacci numbers are typically defined with $a_1=0$ and $a_2=1$ but if we chose $a_2=4$ we would get a very different sequences by applying the relation $a_n=a_{n-1} + a_{n-2}$ because $0+1=1$ while $0+4=4$, a number which is never a Fibonacci number.
However once you establish what those initial conditions are the sequence is entirely formulaic and forced to be what they are by those numbers. In the case you've given we have a single number as an initial condition $a_1=11$. We can now describe every other number in the sequence based on that $a_1$. For example $a_4=a_3+3=(a_2+3)+3=((a_1+3)+3)+3$. At each step I make a substitution of the relation but with the index dropping by one. Our goal is to avoid doing this one step at a time and instead jump the queue and just get the the answer directly. If I can correctly calculate $a_n$ when given $a_{n-k}$ and can just set $n-k=1$ and get the full expression for $a_n$. Since $n-k=1$ implies that $k=n-1$ you make that substitution and it solves the expression in terms of the initial conditions given by $a_1$.
