Question

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$

Answer 1

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{2^n}$

It forms a GP with $a=\frac{1}{2}r=\frac{1/4}{1/2}=1/2$

Sum of n terms of GP $=\frac{a(1-r^n)}{1-r}$

$=\frac{\frac{1}{2}(1-{\left(\frac{1}{2}\right)}^n)}{1-1/2}$

$\therefore$Sum $=1-\frac{1}{2^n}$