How are these 2 sums equal?

Question

I was reading a proof on a formula for the sum of a geometrical progression and these were 2 steps:

$$ = a + x\sum_{0 \le j \le n-1}{ax^j}$$ $$ = a+x\sum_{0 \le j \le n}{ax^j - ax^{n+1}}$$

Apparently, the reason this is true because, according to my book, if $R(j)$ and $S(j)$ are 2 arbitrary relations, then:

$\sum_{R(j)}{a_j} + \sum_{S(j)}{a_j} = \sum_{R(j) \cup S(j)}{a_j} + \sum_{R(j) \cap S(j)}{a_j}$.

However, I'm not sure how that applies here. (Also, the book didn't use the $\cap$ or $\cup$ symbols, it just used "or" and "and" respectively).

Any help would be appreciated.

Answer 1

All that's going on is that $$x\sum_{0\leq j\leq n} ax^j=x\sum_{0\leq j\leq n-1} ax^j+x(ax^n) =x\sum_{0\leq j\leq n-1} ax^j+ax^{n+1} $$ so subtracting over the $ax^{n+1}$ gives you the desired result.

Answer 2

Apply $$ \sum_{0\le j\le n-1}b_j=\sum_{0\le j\le n}b_j-b_n $$ with $b_j=ax^j$.

If you want to see the connection with the notations of your book, then take $R(j):1\le j\le n-1$, $S(j):j=n$ and $a_j=ax^j$.