Đặt: $D=2^x+2^{x+1}+...+2^{x+2015}$
$⇒D=2^x.1+2^x.2+...+2^x.2^{2015}$
$⇒D=2^x.(1+2+...+2^{2015})$
Đặt: $A=1+2+...+2^{2015}$
$⇒2A=2+...+2^{2016}$
$⇒2A-A=-1+(2-2)+...+2^{2016}$
$⇒A=2^{2016}-1$
$⇒D=2^x.(2^{2016}-1)$
Công thức tổng quát:
$2^{n+k}=2^n.2^k$