Sum of powers - how to solve $\sum_{r=1}^{10} 4^r$

Question

How do you calculate sums such as $\displaystyle \sum_{r=1}^{10} 4^r$?

I know how to calculate geometric/arithmetic progressions, but I'm not quite sure how to calculate such 'power' sums using either of those.

Answer 1

For any $x \neq 1$ we know $$x^{n+1}-1 = (x-1)(x^n + x^{n-1} + \ldots + 1).$$ So $$ \sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1} $$ Now since your sum starts at $4^1$ not $4^0$, you need to subtract $1$ in this case to see that $$\sum_{k=1}^{10} 4^k = (4^{11}-1)/(4-1) - 1$$

Answer 2

Take this general formula

$$\sum_{r=p}^q x^r=x^p\frac{x^{q-p+1}-1}{x-1},\quad \forall x\ne1$$