How do break down this addition?

Question

I've been given the following expression:

$2(a + b) + (n + 1)(2a + c) + 2n(2a + d + b) + (a + r)$

And I've been told that it can be simplified to:

$n(6a + 2b + c + 2d) + (5a + 2b + c + r)$

I've tried breaking it down for a while and couldn't find it.

$2a + 2b + 2a(n + 1) + c(n + 1) + n(2a + d + b) + 4a + 2d + 2b + a + r$

$2a + 2b + 2a(n + 1) + c(n + 1) + n(2a + d + b) + 5a + 2d + 2b + r$

$Don't\ know\ how\ to\ go\ from\ here...$

What are the correct order of operations to make it equal the above?

Answer 1

$$2(a+b)+(n+1)(2a+c)+2n(2a+d+b)+(a+r) = $$

$$= 2a+2b+2na+2a+nc+c+4na+2nd+2nb+a+r = $$

$$= (6na+2nb+nc+2nd)+5a+2b+c+r $$

Answer 2

I think there is an error in either the question or the answer, please check again.

Either way, observe the answer carefully. The terms with $n$ are grouped together and the terms without $n$ are grouped together. So when you break it, keep the terms with $n$ together and the other terms together, i.e., break like this:

$ 2(a+b) + n(2a+c) + (2a+c) + n (2a+d+b) + 2(2a+d+b) + (a+r) $

and then group the terms with $n$ together.