Question

For a real number \(x\), \([x]\) denotes the greatest integer less than or equal to \(x\). Then the value of \(\left[ \frac{1}{2} \right] + \left[ \frac{1}{2} + \frac{1}{100} \right] + \left[ \frac{1}{2} + \frac{2}{100} \right] + \dots + \left[ \frac{1}{2} + \frac{99}{100} \right] =\)

• A. \(49\)
• B. \(100\)
• C. \(0\)
• D. \(50\)

Hint:

In floor-function sums, do not compute all terms individually. Instead, find the breakpoint where the floor value changes.

Answer 1

The Correct Option is D

Concept:
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). ip

Step 1: Write the general term.
The terms are of the form: \[ \left[\frac12+\frac{k}{100}\right] \quad \text{for } k=0,1,2,\dots,99 \] ip

Step 2: Decide when the value is \(0\) and when it is \(1\).
Now, \[ \frac12+\frac{k}{100}<1 \quad \text{when } k<50 \] So for \[ k=0,1,2,\dots,49 \] the value of the floor is: \[ 0 \] Also, \[ \frac12+\frac{k}{100}\ge 1 \quad \text{when } k\ge 50 \] So for \[ k=50,51,\dots,99 \] the floor value is: \[ 1 \] ip

Step 3: Count the number of ones.
There are: \[ 50 \] terms equal to \(1\), and \(50\) terms equal to \(0\). Thus total sum is: \[ 50 \] ip Hence, the correct answer is:
\[ \boxed{(D)\ 50} \]