Question

Ajutooooooor!!
Cu cat este egal a, unde a=1+2+2²+2³+2⁴+...+2²⁰¹⁴ ?

Answer 1

2a = 2 + 2^2 + 2^3 + .... + 2^2014 + 2^2015 si
 a = 1+2+2^2+2^3+...+2^214
__________________________ scazi una sub cealalta cele doua egalitati si obtii
 a = 2^2015 - 1.

Bafta!

Answer 2

Folosesti formula generala:

$x^{n+1} -1 = (x-1)(1+x+ x^{2} + x^{3} + ... + x^{n} )$

In cazul tau ai x=2 si n=2014, deci

$2^{2014+1} -1 = (2-1)(1+2+ 2^{2} + 2^{3} + ... + 2^{2014} )$, de unde:

a=$1+2+ 2^{2} + 2^{3} + ... + 2^{2014}$=$2^{2015} -1$