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

Step 1: Understanding the Concept:
We evaluate terms in the sum to find when they transition from \(0\) to \(1\).
Step 2: Key Formula or Approach:
The sum is \(\sum_{k=0}^{99} [1/2 + k/100]\).
A term \([1/2 + k/100] = 1\) if \(1/2 + k/100 \ge 1 \implies k/100 \ge 1/2 \implies k \ge 50\).
Step 3: Detailed Explanation:
For \(k = 0, 1, \dots, 49\), the values are \(\lt 1\), so their floor is \(0\). Total sum from these terms \(= 0\).
For \(k = 50, 51, \dots, 99\), the values are \(\ge 1\) but \(\lt 2\), so their floor is \(1\).
Number of terms from \(50\) to \(99\) is \(99 - 50 + 1 = 50\).
Total Sum \(= 50 \times 1 = 50\).
Step 4: Final Answer:
The sum is \(50\).