Key Concepts and Formulas

• Greatest Integer Function (GIF) Definition: For any real number $x$, $[x]$ denotes the greatest integer less than or equal to $x$. This implies that if $n$ is an integer, then $n \le x < n+1 \iff [x] = n$.
• Inequality Manipulation: When solving inequalities, multiplying or dividing by a negative number reverses the direction of the inequality sign.
• Arithmetic Series Sum: The sum of an arithmetic series with $N$ terms, first term $a_1$, and common difference $d$ is $S_N = \frac{N}{2}(2a_1 + (N-1)d)$. Alternatively, if the first term is $a$ and the last term is $l$, the sum is $S_N = \frac{N}{2}(a+l)$.

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Step-by-Step Solution

Step 1: Analyze the General Term and the Series Structure

The given series is $S = \left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right]$. We can express the general term of the series as $T_k = \left[ - \frac{1}{3} - \frac{k}{100} \right]$, where $k$ ranges from $0$ to $99$. The total number of terms in the series is $99 - 0 + 1 = 100$. Let $X_k = - \frac{1}{3} - \frac{k}{100}$. We need to determine the integer value of $[X_k]$ for each $k$.

Step 2: Determine the Range of Values for the General Term

The value of $[X_k]$ depends on the interval in which $X_k$ lies. We know that $\frac{1}{3} = 0.333...$. Let's analyze the possible integer values for $[X_k]$.

• Consider the first term ($k=0$): $X_0 = - \frac{1}{3} - \frac{0}{100} = - \frac{1}{3}$. Since $-1 \le - \frac{1}{3} < 0$, we have $\left[ - \frac{1}{3} \right] = -1$.

• Determine when $[X_k]$ changes value: The value of $[X_k]$ will change when $X_k$ crosses an integer. The smallest possible integer value for $[X_k]$ is $-1$ (as seen for $k=0$), and as $k$ increases, $X_k$ becomes more negative. We need to find the values of $k$ for which $X_k$ falls into different integer intervals.

  Case 1: $[X_k] = -1$ This occurs when $-1 \le X_k < 0$. Substituting $X_k$: $-1 \le - \frac{1}{3} - \frac{k}{100} < 0$ Add $\frac{1}{3}$ to all parts: $-1 + \frac{1}{3} \le - \frac{k}{100} < 0 + \frac{1}{3}$ $- \frac{2}{3} \le - \frac{k}{100} < \frac{1}{3}$ Multiply by $-1$ and reverse the inequalities: $- \frac{1}{3} < \frac{k}{100} \le \frac{2}{3}$ Multiply by $100$: $- \frac{100}{3} < k \le \frac{200}{3}$ $-33.33... < k \le 66.66...$ Since $k$ starts from $0$, the integers $k$ in this range are $0, 1, 2, \dots, 66$. The number of terms for which $[X_k] = -1$ is $66 - 0 + 1 = 67$.

  Case 2: $[X_k] = -2$ This occurs when $-2 \le X_k < -1$. Substituting $X_k$: $-2 \le - \frac{1}{3} - \frac{k}{100} < -1$ Add $\frac{1}{3}$ to all parts: $-2 + \frac{1}{3} \le - \frac{k}{100} < -1 + \frac{1}{3}$ $- \frac{5}{3} \le - \frac{k}{100} < - \frac{2}{3}$ Multiply by $-1$ and reverse the inequalities: $\frac{2}{3} < \frac{k}{100} \le \frac{5}{3}$ Multiply by $100$: $\frac{200}{3} < k \le \frac{500}{3}$ $66.66... < k \le 166.66...$ The series terms are for $k$ from $0$ to $99$. The integers $k$ that satisfy this condition and are within the series' range are $67, 68, \dots, 99$. The number of terms for which $[X_k] = -2$ is $99 - 67 + 1 = 33$.

  Case 3: $[X_k] = -3$ (and smaller integers) For $[X_k]$ to be $-3$, we would need $-3 \le X_k < -2$. $-3 \le - \frac{1}{3} - \frac{k}{100} < -2$ $- \frac{8}{3} \le - \frac{k}{100} < - \frac{5}{3}$ $\frac{5}{3} < \frac{k}{100} \le \frac{8}{3}$ $\frac{500}{3} < k \le \frac{800}{3}$ $166.66... < k \le 266.66...$ Since the maximum value of $k$ in the series is $99$, there are no terms in the given series for which $[X_k]$ is $-3$ or any smaller integer.

Step 3: Calculate the Sum of Terms in Each Case

• For $k = 0, 1, \dots, 66$ (67 terms), the value of each term is $-1$. The sum of these terms is $67 \times (-1) = -67$.

• For $k = 67, 68, \dots, 99$ (33 terms), the value of each term is $-2$. The sum of these terms is $33 \times (-2) = -66$.

Step 4: Calculate the Total Sum of the Series

The total sum of the series is the sum of the partial sums from the different cases: $S = (\text{Sum of terms where } [X_k] = -1) + (\text{Sum of terms where } [X_k] = -2)$ $S = (-67) + (-66)$ $S = -133$

Common Mistakes & Tips

• GIF with Negative Numbers: Be extremely careful when evaluating the GIF for negative numbers. For example, $[-0.01] = -1$, not $0$. The value is always the greatest integer less than or equal to the number.
• Inequality Reversal: Remember to reverse the inequality signs when multiplying or dividing by a negative number. This is a frequent source of errors in GIF problems.
• Counting Terms: Ensure you correctly count the number of terms in each range. For a range from $a$ to $b$ inclusive, the count is $b - a + 1$.

Summary

The problem involves calculating the sum of a series where each term is the greatest integer of a specific expression. We first defined the general term and then analyzed the ranges of the index $k$ for which the greatest integer function evaluates to specific integer values. By identifying that the terms take the value $-1$ for $67$ values of $k$ and $-2$ for the remaining $33$ values of $k$, we calculated the sum of these two groups of terms separately and then added them to find the total sum of the series.

The final answer is $\boxed{\text{-133}}$.