Question

If for a real number $x,\left[x\right]$ denotes the greatest integer less than or equal to $x,$ then for any $n\in N$, $\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\left[\frac{n+4}{8}\right]+\left[\frac{n+8}{16}\right]+...=$

• A. $n$
• B. $n-1$
• C. $n+1$
• D. $n+2$

Answer 1

The correct option is A $n$

For any $x\in R$ we have

$\left[x\right]=\left[\frac{x}{2}\right]+\left[\frac{x+1}{2}\right]$

$\Rightarrow \left[n\right]=\left[\frac{n+1}{2}\right]+\left[\frac{n}{2}\right]$ ....... $\left(i\right)$

$\Rightarrow n=\left[\frac{n+1}{2}\right]+\left[\frac{n}{2}\right]$

$\Rightarrow n=\left[\frac{n+1}{2}\right]+\left[\frac{n}{4}\right]+\left[\frac{\frac{n}{2}+1}{2}\right]$ $[$using Eq, $\left(i\right)]$

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

$\Rightarrow n=\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\left[\frac{n}{8}\right]+\left[\frac{\frac{n}{4}+1}{2}\right]$ $[$using Eq. $\left(i\right)]$

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

Continuing in this manner, we have

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