Question

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

Answer 1

Use the property of greatest integer function, $\left[ x \right] = \left[ {\dfrac{x}{2}} \right] + \left[ {\dfrac{{x + 1}}{2}} \right]$ and expand the value of $\left[ n \right]$, where $n$ is a natural number. Write $n$ as $\left[ {\dfrac{n}{2}} \right] + \left[ {\dfrac{{n + 1}}{2}} \right]$. Then, expand $\left[ {\dfrac{n}{2}} \right]$ by the same property and continue like this to get an equivalent expression.

Complete step by step solution:
We know that if $\left[ x \right]$ is a greatest integer less than or equal to $x$, then we have $\left[ x \right] = \left[ {\dfrac{x}{2}} \right] + \left[ {\dfrac{{x + 1}}{2}} \right]$
Let us write the same definition for $\left[ n \right]$
$\left[ n \right] = \left[ {\dfrac{n}{2}} \right] + \left[ {\dfrac{{n + 1}}{2}} \right]$
Since, $n \in N$, we can write this as,
$n = \left[ {\dfrac{{n + 1}}{2}} \right] + \left[ {\dfrac{n}{2}} \right]$ eqn. (1)
Similarly, we can write,
$\left[ {\dfrac{n}{2}} \right] = \left[ {\dfrac{n}{4}} \right] + \left[ {\dfrac{{\dfrac{n}{2} + 1}}{2}} \right] \\\ \Rightarrow \left[ {\dfrac{n}{2}} \right] = \left[ {\dfrac{n}{4}} \right] + \left[ {\dfrac{{n + 2}}{4}} \right] \\\$
When we substitute the value of $\left[ {\dfrac{n}{2}} \right]$ in equation (1), we will get,
$n = \left[ {\dfrac{{n + 1}}{2}} \right] + \left[ {\dfrac{{n + 2}}{4}} \right] + \left[ {\dfrac{n}{4}} \right]$
We will now similarly, expand $\left[ {\dfrac{n}{4}} \right]$
$n = \left[ {\dfrac{{n + 1}}{2}} \right] + \left[ {\dfrac{{n + 2}}{4}} \right] + \left[ {\dfrac{n}{8}} \right] + \left[ {\dfrac{{\dfrac{n}{4} + 1}}{2}} \right] \\\ \Rightarrow n = \left[ {\dfrac{{n + 1}}{2}} \right] + \left[ {\dfrac{{n + 2}}{4}} \right] + \left[ {\dfrac{{n + 4}}{8}} \right] + \left[ {\dfrac{n}{8}} \right] \\\$
And here, we can observe that if we continue like this, we will have,
$n = \left[ {\dfrac{{n + 1}}{2}} \right] + \left[ {\dfrac{{n + 2}}{4}} \right] + \left[ {\dfrac{{n + 4}}{8}} \right] + \left[ {\dfrac{{n + 8}}{{16}}} \right] + .......$
Thus, the given expression is equal to $n$.

Hence, option A is the correct option.

Note:
We represent the greatest integer function as $\left[ x \right]$. If a number is such that $\left[ {4.2} \right]$, then the value of the number is 4. Similarly, if we have $\left[ 4 \right]$, then it is equal to 4. Greatest integer function is also known as step-function.