Expanding, The Necessary Evil

Nice problem. This question has been heavily veiled with complicated looking symbols.

Let's solve the 2 innermost brackets.

$(x - 1)(\displaystyle \sum_{i = 0}^j x^i) = (x - 1)(1 + x + x^2 + \ldots + x^j)$

$= x^{j +1} - 1$

Whoa! How did I do that?! I just simplified it. You can experiment with known values of $n$ and find it always total to this. Following this principle, we continue with the question.

$\dfrac {(x^{j + 1} - 1) + 1}{x}$

$\frac {x^{j + 1}}{x}$

$x^j$

Next stage.

$(x - 1)(\displaystyle \sum_{j = 1}^{n + 1} x^j) = (x - 1)(x + x^2 + \ldots + x^{n + 1})$

$= x^{n + 2} - x$

Final stage. This part is relatively simple.

$\frac {x^{n + 2} - x}{x} = \frac {x(x^{n + 1} - 1)}{x}$

$= x^{n + 1} - 1$

Magic!! :D

I think the author is trying to tell us to find

$\dfrac{(x-1) \displaystyle\sum_{n=1}^{ m+1}\left( \dfrac{(x-1)\left(\displaystyle\sum_{i=0}^{n} x^i\right)+1}{x}\right)}{x}$

if so, the answer would be $x^{m+1}-1$

You can write $\displaystyle\sum_{i=0}^{n}x^i$ as $\dfrac{1-x^{n+1}}{1-x}$

So, $(x-1)\left(\displaystyle\sum_{i=0}^{n}x^i\right)$ will become $x^{n+1}-1$

and $\dfrac{(x-1)\left(\displaystyle\sum_{i=0}^{n} x^i\right)+1}{x}$ becomes $x^n$

Now, the expression looks like $\dfrac{(x-1)\left(\displaystyle\sum_{n=1}^{m+1}x^n\right)}{x}$

which can be further simplified to $x^{m+1}-1$ .