Chapter 7: Problem 75
For \(x \in R\), let \([x]\) denotes the greatest integer \(\leq x\), then the value of \(\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]\) \(+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]\) is (A) \(-100\) (B) \(-123\) (C) \(-135\) (D) \(-153\)
Short Answer
Step by step solution
Understanding the Problem
Analyzing the Greatest Integer Function
Determining Each Term
Counting and Summing Values
Calculating the Total Sum
Adjusting for Miscalculation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Greatest Integer Calculations
- The function ignores the decimal part and rounds down to the nearest integer, even if the number is negative.
- This is particularly useful in sequences where fractional parts are included but only integer parts are needed for calculations.
Integer Sequences
In the given problem, we examined a sequence created by evaluating the expression \(-\frac{1}{3} - \frac{k}{100}\) for \(k\) from 0 to 99. Here's how this sequence plays out:
- The sequence started by determining the integer value of each term using the greatest integer function, initially yielding values like \([ - \frac{1}{3} ] = -1\).
- As \(k\) increased, the expression \(-\frac{1}{3} - \frac{k}{100}\) decreased. Beyond a certain point, it required adjusting to the next lower integer, i.e., from -1 to -2.
Mathematical Problem-Solving
- Breaking the expression \( -\frac{1}{3} - \frac{k}{100} \) into a sequence allowed for focused calculation of each individual term using the greatest integer function.
- Identifying the point where the sequence shifted its integer value involved understanding both progression and decrement within the real numbers.
- Counting terms accurately was essential in computing the sum. Initially, we calculated 67 occurrences for \([x] = -1\) and 33 for \([x] = -2\), highlighting how counting contributes heavily to finding the correct answer.