Solve Recurrence Relation
It's probably easier solving this using the Master Theorem. 

T(n) = 6T(n/6) + 2n + 3




So in this case A=6, B=6, and D=1 because f(n) = 2n + 3 and 2n^1 + 3 = 2n + 3 

We of course assume the base case is a constant such that T(1) = C

So we can easily see that the answer for this is 

  T
  (
  n
  )
  =
  &#x0398;
  (
  n log n
  )
, since A is equal to B to the power of D.

We can see that is true just by plugging in the values

= A = B^D  

= 6 = 6^1

= 2 > 1 (which is always true)

This guy on YouTube has a very nice video explaining how to solve recurrences using the Master theorem, I will leave the link here.