Rewriting summation formula I just watched a tutorial on recurrence by substitution. In the tutorial, it mentioned about rewriting
$\sum\limits_{i=1}^\mathbb{k}{2^i}$
as (2k+1 - 2). My question is can I generalize it as xlimit + 1 - x where x is the base. 

Let
$$a(n)=b^n$$
for some $b$; then
$$\Delta a(n)=a(n+1)-a(n)=b^{n+1}-b^n=b^n(b-1);$$
divide both sides by $b-1$:
$$b^n=\frac{\Delta a(n)}{b-1};$$
substitute back for $a(n)$:
$$a(n)=b^n=\frac{\Delta (b^n)}{b-1}={\large\Delta}\left(\frac{b^n}{b-1}\right).$$
Evaluate
$$\sum\limits_{k=1}^na(k).$$
Apply the fundamental theorem of discrete calculus:
$$\sum\limits_{k=1}^na(k)=\bigg({\large\Delta}^{-1}a(k)\bigg)\Bigg|_{k=1}^{n+1}=\left(\frac{b^k}{b-1}\right)\Bigg|_{k=1}^{n+1}=\frac{b^{k+1}}{b-1}-\frac{b}{b-1};$$
write out the answer:
$$\therefore\sum\limits_{k=1}^nb^k=\frac{b^{k+1}-b}{b-1}.$$
