Question

The remainder, if $1+2+2^2+2^3+....+2^{1999}$ is divided by $5$ is?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$1+2+2^2+.....+2^{1999}$ is in geometric progression

So we get $1+2+2^2+.....+2^{1999}=\frac{2^{2000}-1}{2-1}=2^{2000}-1$

$2^{2000}-1$ can be written as ${16}^{500}-1={(15+1)}^{500}-1={{}^{500}C}_0\times {15}^{500}+{{}^{500}C}_1\times {15}^{499}+{{}^{500}C}_2\times {15}^{498}+.....+{{}^{500}C}_{500}\times {15}^0-1=15\times constant$

Therefore the given series is divisible by $5$

So the remainder is zero

Therefore option $A$ is correct