For x ∈ R, let [x] denote the greatest integer ≤ x, then the sum of the series [1/3] + [1/3 + 1/100]

Q: Wait, the question clearly has positive signs inside the GIF. Why are we allowed to just change it to negative?

In the context of JEE Advanced, when you encounter a mathematical impossibility—like a sum of positive terms resulting in a negative value—it is almost certainly a typo. Rather than panicking, look at the options. The options are negative, which confirms the intended logic. As an engineer, you must learn to identify when the problem statement is flawed and pivot to the logical intent.

Q: Why is the GIF of $-1.003$ equal to $-2$ and not $-1$?

This is a classic trap! Remember the definition: $[x]$ is the greatest integer less than or equal to $x$. On the number line, $-1$ is to the right of $-1.003$, meaning $-1 > -1.003$. Therefore, $-1$ cannot be the GIF. The integer to the left of $-1.003$ is $-2$. Always visualize the number line when dealing with negative GIF values.

Q: How do I know where to split the summation? Is there a formula?

There isn't a single formula, but there is a method. You are looking for the transition points where the argument of the GIF crosses an integer. You set the argument equal to an integer $k$ and solve for $r$. In this problem, the argument $-\frac{1}{3} - \frac{r}{100}$ starts at $-0.333$ (when $r=0$) and decreases. It will hit $-1$ at $r=66.66$. Since $r$ must be an integer, the transition happens between $66$ and $67$. That is your split point.

Question

For $x \in R$, let $[x]$ denote the greatest integer $\le x$, then the sum of the series $[\frac{1}{3}] + [\frac{1}{3} + \frac{1}{100}] + [\frac{1}{3} + \frac{2}{100}] + \dots + [\frac{1}{3} + \frac{99}{100}]$ is

Options:

(A) $-153$

(B) $-133$

(C) $-131$

(D) $-135$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, today we are going to dissect a problem that serves as a perfect lesson in mathematical intuition. You are looking at a series:

$$\sum_{r=0}^{99} \left[ \frac{1}{3} + \frac{r}{100} ight]$$

You might be tempted to start calculating immediately, but stop. Look at the options: $-153, -133, -131, -135$.

They are all negative. If you sum positive numbers, you get a positive result. This is a classic JEE moment where the question has a typo. We must assume the intended series is:

$$S = \sum_{r=0}^{99} \left[ -\frac{1}{3} - \frac{r}{100} ight]$$

Let us proceed with this corrected logic.

### The Philosophy of the Step Function

The Greatest Integer Function (GIF) is not just a bracket; it is a staircase. It stays flat, holding a constant value, until it hits an integer, at which point it drops instantly to the next level.

For positive numbers, this is intuitive. For negative numbers, it is where most students stumble. Remember, $[x]$ is the largest integer less than or equal to $x$.

If $x = -0.333$, the integers less than it are $-1, -2, -3, \dots$. The largest of these is $-1$. So, $[-0.333] = -1$. This is our starting point.

### The Hunt for the Transition

We need to know when this staircase drops. Our general term is $T_r = [-\frac{1}{3} - \frac{r}{100}]$. As $r$ increases, the value inside the bracket becomes more negative, moving left on the number line.

We need to find when this value crosses $-1$. We set the argument equal to $-1$:

$$-\frac{1}{3} - \frac{r}{100} = -1$$

Solving this, we get:

$$\frac{r}{100} = 1 - \frac{1}{3} \Rightarrow \frac{r}{100} = \frac{2}{3}$$

Thus, $r = \frac{200}{3} \approx 66.66$. This is our critical boundary.

For any integer $r \le 66$, the value is still greater than $-1$, so the GIF remains $-1$. For any integer $r \ge 67$, the value drops below $-1$, so the GIF becomes $-2$.

### The Final Summation

Now, we partition the series. The first interval is $r=0$ to $r=66$. The number of terms is $66 - 0 + 1 = 67$.

Each of these terms contributes $-1$ to the sum. So, the first part is:

$$67 	imes (-1) = -67$$

The second interval is $r=67$ to $r=99$. The number of terms is $99 - 67 + 1 = 33$. Each of these terms contributes $-2$ to the sum.

So, the second part is:

$$33 	imes (-2) = -66$$

Adding these together, we get:

$$S = -67 - 66 = -133$$

We have navigated the typo, understood the behavior of the GIF, and arrived at the correct answer of **$-133$**. This is the essence of JEE Advanced—not just calculation, but the ability to see the structure beneath the symbols.

Animated Solution for Mathematics - Functions: For $x \in R$ , let $[x]$ denote the greatest integer $\leq x$ , then the sum of the series $[\frac{1}{3}] + [\frac{1}{3} + \frac{1}{100}] + [\frac{1}{3} + \frac{2}{100}] + \dots + [\frac{1}{3} + \frac{99}{100}]$ is

Select Answer:

Visualized Solution

Analyzing the Series and Options

Greatest Integer Function (GIF) Basics

Evaluating the First Term ( $r = 0$ )

Finding the Transition Point

Solving for $r$

The First Interval ( $r = 0$ to $66$ )

Summing the First Interval

The Second Interval ( $r = 67$ to $99$ )

Summing the Second Interval

Final Calculation

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Philosophy of the Step Function

The Hunt for the Transition

The Final Summation

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Visualized Solution

Analyzing the Series and Options

Greatest Integer Function (GIF) Basics

Evaluating the First Term ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=0)

Finding the Transition Point

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

The First Interval ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=0 to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 66)

Summing the First Interval

The Second Interval ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=67 to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 99)