Sequences & Series - the sum to $n$ terms of a progression $$(a + b) + (a^2 + ab\ +\ b^2) + (a^3 + a^2b + ab^2 + b^3) +\cdots$$
It is clear that the objective is to find an expression for the sum of this sequence to 'n' terms. I discerned that it is not an arithmetic, geometric, harmonic or arithmetico-geometric sequence, in which case I could have applied the necessary formula to obtain my answer.
Therefore, I was able to find a general expression for the nth term of this sequence which is as follows:
$$t_i\ =\ (a\ +\ b)^i\ -\ (i\ -\ 1)(ab)(a\ +\ b)^{i\ -\ 2}$$
Then I tried to find the summation of this expression from $i\ =\ 1$ to $i\ =\ n$.
$$ \sum_{i=1}^{n} [(a\ +\ b)^i\ -\ (i\ -\ 1)(ab)(a\ +\ b)^{i - 2}]$$
Putting $p\ =\ (a\ +\ b)$ and $q\ =\ (ab)$.
The expression becomes:
$$\sum_{i=1}^{n} [p^i\ -\ (i\ -\ 1)q(p)^{i\ -\ 2}]$$
$$\sum_{i=1}^{n} p^i\ -\ q\sum_{i=1}^{n} (ip^{i\ -\ 2}\ -\ p^{i\ -\ 2})$$
This becomes:
$$ p\left[\frac{p^i - 1}{p - 1}\right] -\left(q\frac{(i^2 + i)}{2}\right) \left(k\left[\frac{k^{i - 2}}{k - 1}\right] - k\left[\frac{k^{i - 2} - 1}{k - 1}\right]\right)$$
As I am unable to simplify this expression completely, I am unable to compare my answer with the final answer given in my Algebra textbook. The final answer, according to the book, comes out to be:
$$\large \frac{1}{(a - b)} \frac{a^2(1 - a^n)}{1 - b} - \frac{1}{(a - b)} \frac{b^2(1 - b^n)}{1 - b}$$
Is my approach to this question correct? If so, does it eventually give me this final answer? Otherwise, can someone tell me how to solve this problem? 

 A: The series can be rewritten (provided $a\neq b$) as
$$
\frac{a^2-b^2}{a-b}+\frac{a^3-b^3}{a-b}+\frac{a^4-b^4}{a-b}+\dots,
$$
so it's the difference of two geometric series.
A: Consider the case $i=5$. Then 
$$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+a^5,$$
 and
$$4ab(a+b)^3=4a^4b+12a^3b^2+12a^2b^3+4ab^4.$$
The difference is not the desired $a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$. 
Remark: So the initial representation was not correct. But we show how the summation you were attempting could have been achieved.  Suppose we want to know $\sum_1^n (i-1)ab(a+b)^{i-2}$. Let $x=a+b$. So we want $\sum_1^n (i-1)x^{i-2}$. 
Note that the sum $F(x)=\sum_1^{n} x^{i-1}$ is a geometric series, with,  (if $x\ne 1$) a relatively simple expression for the sum. Then the sum   $\sum_1^n (i-1)x^{i-2}$ is just $F'(x)$, the derivative of $F(x)$. 
There are more elementary ways to do the calculation. Consider for example the sum $1+2x+3x^2+4x^3+5x^4+6x^5+7x^6$. Call this sum $G(x)$. Then $G(x)-xG(x)=(1+x+\cdots +x^6)-7x^7$. The geometric series $1+x+\cdots+x^6$ has a relatively simple sum, and now we can solve for $G(x)$. The same method works in general.
