# Calculate A, where A = 2 + 2xx2 + 2xx2xx2 + 2xx2xx2xx2 + ... + 2xx2 ... 2xx2 (the last number has 2013 digits 2)?

May 3, 2018

$A = \frac{2 \left(1 - {2}^{2013}\right)}{1 - 2}$

#### Explanation:

This is represented by

${\sum}_{n = 1}^{2013} {2}^{n}$

This is a geometric series of common ratio $2$ and first term $2$.

${S}_{n} = \frac{2 \left(1 - {2}^{2013}\right)}{1 - 2}$

This is very large so I won't even try evaluating.

Hopefully this helps!