$\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+............$to n terms =

We have,

$\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+............$to n terms

$=\frac{2^2-1}{2^2}+\frac{2^4-1}{2^4}+\frac{2^6-1}{2^6}+......$to n terms

$=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^4}\right)+\left(1-\frac{1}{2^6}\right)+........$to n terms

$=\mathrm{n}-\left\{\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+........\mathrm{to} \mathrm{n} \mathrm{terms}\right\}$

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

$=\mathrm{n}-\frac{1}{3}\left(1-4^{-\mathrm{n}}\right)=\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}$