Тогда так:
2^x(1+(1/2)+(1/4)+...+(1/2^(2000)) > 2^(2017)
В скобках сумма геометрической прогрессии b_(1)=1; q=(1/2); n=2001
Формула
S_(n)=b_(1)(1-q^n)/(1-q)
S_(2001)=(1-(1/2)^(2000))/(1-(1/2))=2*(1-(1/2)^(2000))
2^(x+1)*(1-(1/2)^(2000) > 2^(2017)
2^(x+1) > 2^(x+1)*(1-(1/2)^(2000) > 2^(2017)
x+1 > 2017
x > 2016
n=2017