Why is $\sum_{i=1}^6 2^i = 2^7-2$?

Question

Why is $$\sum_{i=1}^6 2^i = 2^7-2$$

Answer 1

$$\sum_{i=1}^6 2^i =2+2^2+2^3+2^4+2^5+2^6$$ $$= 2(1+2+2^2+..+2^5) = 2{2^6-1\over 2-1} =2^7-2$$

We have generaly $$1+x+x^2+...+x^n = {x^{n+1}-1\over x-1}$$

Answer 2

We use Geometric series (see https://en.wikipedia.org/wiki/Geometric_series). That is, we have $$ \sum_{i=0}^k n^i = \frac{n^{k+1} - 1}{n - 1}.$$ So rewriting your series, we have $$ \sum_{i=1}^6 2^i = 2 \sum_{i=0}^5 2^i = 2 \left(\frac{2^6 - 1}{2 -1} \right) = 2^7 - 2.$$

Answer 3

In context, binary arithmetic:

$\sum_{i=1}^{6}2^i= 1111110.$

$\sum_{i=1}^{6}2^i +2=$

$1111110+10=$

$10000000 =2^7.$

Answer 4

The easiest way to see it is to note that the series is telescoping. Indeed, $$ \begin{align} \sum_{i=1}^6 2 ^i&=\sum_{i=1}^6 (2 ^{i+1}-2^i)\\&=(2^2-2^1)+(2^3-2^2)+ (2^4-2^3)+ (2^5-2^4)+ (2^6-2^5)+ (2^7-2^6)\\ &=2^7-2. \end{align} $$