The question I am trying to solve is part b) of this question:
In the marking instructions the solution is all completely clear except from the substitution $2nI_n=(2n-1)I_n$, I simply cannot see where it comes from: 
Can anyone indicate why does this is true and where it comes from, I've spent quite a lot of time playing with the equations and I've been completely unsuccessful so far.
Your first equation reads $$I_n = \frac{1}{2^n} + 2nI_n - 2nI_{n+1}$$ Subtract $I_n$ from both sides and then add $2nI_{n+1}$ to both sides. This becomes: $$2nI_{n+1} = \frac{1}{2^n} + (2n-1) I_n$$ as written in your book. There's no substitution involved, just a subtraction and an addition to both sides of an equation.
PS: This is why you should put $\iff$ signs between equivalent equations... At the very least it makes clear to everyone that it's a new equation, not the continuation of the preceding string of equal signs.