Problem 106 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

For \(x \in \mathrm{R}\), let \([x]\) denote the greatest integer \(\leq x\), then the sum of the series \(\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\cdots+\left[-\frac{1}{3}-\frac{99}{100}\right]\) is [April 12, 2019 (I)] (a) \(-153\) (b) \(-133\) (c) \(-131\) (d) \(-135\)

Step-by-Step Solution

Answer

(b) -133

Step 1: Understand the Greatest Integer Function

The function \([x]\) represents the greatest integer less than or equal to \(x\). For instance, \([-\frac{1}{3}] = -1\) because \(-1\) is the greatest integer less than or equal to \(-\frac{1}{3}\). We need to address how this function behaves for each term in the series.

Step 2: Analyze the Series Terms

Starting with the first term: \([-\frac{1}{3}]= -1\). We observe that this is because \(-0.3333...\) rounds down to \(-1\). For the second term: \([-\frac{1}{3} - \frac{1}{100}] = [-0.3433...] = -1\). This pattern will shift from \(-1\) to \(-2\) when the expression inside \([\cdot]\) crosses a boundary that rounds beyond \(-1\).

Step 3: Calculate the Transition Point

To find when the terms switch from rounding to \(-1\) to rounding to \(-2\), solve \(-\frac{1}{3} - \frac{n}{100} < -1\), which simplifies to \(-\frac{1}{3} < -1 + \frac{n}{100}\). Solving this, we get \(\frac{2}{3} < \frac{n}{100}\), which implies \(66.67 < n\). Therefore, when \(n = 67\), the term becomes \([-2]\).

Step 4: Sum the Values from Each Transition

Terms from \(n = 0\) to \(n = 66\) evaluate to \([-1]\), totaling \(-1 \times 67 = -67\). Terms from \(n = 67\) to \(n = 99\) evaluate to \([-2]\), totaling \(-2 \times 33 = -66\). Add these partial sums to determine the total sum of the series.

Step 5: Calculate the Total Sum

Add the sums from step 4: \(-67 + (-66) = -133\). Therefore, the sum of the series is \(-133\).

Key Concepts

Series Sum

When examining a series sum, we're basically adding up a list of numbers. In this exercise, we consider a series that involves applying the greatest integer function to each term. This series sum is calculated as:

• Evaluate each term based on its specified formula.
• Apply the appropriate function—in this case, the greatest integer.
• Add up all the results.

This might sound simple, but the complexity arises in understanding how the greatest integer function affects each term. A careful analysis reveals that the series split into two segments: one segment where the terms evaluate to e (-1) and another where they evaluate to (-2). By finding where this transition occurs, we compute each partial sum separately before adding them together. The final result gives you the complete sum of the original series. Understanding a series sum is critical because it reinforces concepts about summation and introduces the effects of special functions on series computations.

Floor Function

The floor function, denoted by For the greatest integer function, often described with the floor notation is a mathematical operation that rounds a real number down to the nearest integer. In this exercise, it was essential to determine how each term in the series evaluated using this function. For example, We know that for any real number \(x\), the floor of \(-\frac{1}{3}\) is \([-1]\).Understanding the floor function helps in situations like these because it tells you precisely how each term will round. This rounding is crucial when these terms are summed in a series. For this problem, each term slightly decreases until enough changes accumulate that it shifts the greatest integer it rounds to.This function is especially useful for working through problems that require precise control over how numbers adjust, especially when dealing with variations of real numbers in series or sequences.

Integer Properties

Integer properties are fundamental to understanding many mathematical concepts, including the behavior of the greatest integer function in this exercise. Here are some key ideas:

• Integers do not include fractions or decimals.
• The behavior of the greatest integer function depends on rounding down to the nearest whole number.
• Sequences of operations on integers, especially addition and multiplication, follow consistent laws like distributive, associative, and commutative properties.

In the context of this problem, integer properties play a significant role when determining how the series terms transition points affected their floor values. Recognizing when many small changes will compose to effect a change in an integer result is critical. This type of exercise helps strengthen your grasp of how operations affect integers, and how you can predict these changes in the correct context.