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

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Answer 1 · 2018-05-03T15:55:25

$A= (2(1 - 2^2013))/(1 - 2)$

This is represented by

$sum_(n = 1)^2013 2^n$

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

$S_n = (2(1 - 2^2013))/(1 - 2)$

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

Hopefully this helps!

Calculate AA, where A = 2 + 2xx2 + 2xx2xx2 + 2xx2xx2xx2 + ... + 2xx2 ... 2xx2A = 2 + 2xx2 + 2xx2xx2 + 2xx2xx2xx2 + ... + 2xx2 ... 2xx2 (the last number has 2013 digits 2)?