Question

$2^{100}-2^{99}-2^{98}\dots \dots -2^2-2^1-2^0=$

• A. 1
• B. 2
• C. $2^{50}$

Answer 1

The correct option is A 1

Given expression $2^{100}-2^{99}-2^{98}\dots \dots -2^2-2^1-2^0$
$=2^{99}(2-1)-2^{98}-2^{97}\dots \dots 2^2-2^1-2^0$
$=2^{99}-2^{98}-2^{97}\dots \dots -2^2-2^1-2^0$
$=2^{98}(2-1)-2^{97}\dots \dots -2^2-2^1-2^0$
$=2^{98}-2^{97}\dots \dots -2^2-2^1-2^0$
$=2^{97}-2^{96}\dots \dots -2^2-2^1-2^0$
Continuing this process until the final step, we get
$=2^3-2^2-2^1-2^0$
$=2^2-2^1-2^0$
$=2^1-2^0$
$=2-1=1$
Hence, the value of the given expression is$=1$