Mathematics - Sequence & Series Question with Solution | TestHub

Consider $\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\dots$ upto $\mathrm{n}$ terms $=\frac{2^2-1}{2^2}+\frac{2^4-1}{2^4}+\frac{2^6-1}{2^6}$ upto n term $=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^4}\right)+\left(1-\frac{1}{2^6}\right)+\dots$ upton terms $=(1+1+1+\dots .$ upto $\mathrm{n}$ terms $)$ $-(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots$ upto n terms $)$ $=\mathrm{n}-\frac{1}{2^2}\left[\frac{1-{\left(\frac{1}{2^2}\right)}^{\mathrm{n}}}{1\frac{1}{2^2}}\right]=\mathrm{n}-\frac{1}{3}\left(1-4^{-\mathrm{n}}\right)$ $=\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}$