If for a real number x, [x] denotes the greatest integer less than or equal to x, then for any n ∈ N

$\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}+2}{4}\right]+\left[\frac{\mathrm{n}+4}{8}\right]+\left[\frac{\mathrm{n}+8}{16}\right]+\dots =$

For any x ∈ R, we have

$\left[\mathrm{x}\right]=\left[\frac{\mathrm{x}}{2}\right]+\left[\frac{\mathrm{x}+1}{2}\right]$  …….(1)

$\begin{array}{l}\Rightarrow \left[\mathrm{n}\right]=\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}}{2}\right]\\ \Rightarrow \mathrm{n}=\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}}{2}\right]\\ =\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}}{4}\right]+\left[\frac{\frac{\mathrm{n}}{2}+1}{2}\right] [\mathrm{using} (1\left)\right]\\ =\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}+2}{4}\right]+\left[\frac{\mathrm{n}}{4}\right]\\ =\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}+2}{4}\right]+\left[\frac{\mathrm{n}}{8}\right]+\left[\frac{\frac{\mathrm{n}}{4}+1}{2}\right]\end{array}$

[Using (1)]

    $=\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}+2}{4}\right]+\left[\frac{\mathrm{n}+4}{8}\right]+\left[\frac{\mathrm{n}}{8}\right]$

Continuing in this manner, we have,

$\left[\frac{\mathrm{n}+1}{2}\right]+\left[\frac{\mathrm{n}+2}{4}\right]+\left[\frac{\mathrm{n}+4}{8}\right]+\dots =\mathrm{n}$