Question

For x ∈ R, let [x] denote the greatest integer ≥ x, then the sum of the series $\left[-\frac{1}{3}\right] + \left[-\frac{1}{3}-\frac{1}{100}\right]+........+\left[-\frac{1}{3}-\frac{99}{100}\right]$ is:

• A. -153
• B. -133
• C. -135
• D. -131

Answer 1

Answer: (B)

-133

Answer 2

The given series is $S = \left[-\frac{1}{3}\right] + \left[-\frac{1}{3}-\frac{1}{100}\right]+........+\left[-\frac{1}{3}-\frac{99}{100}\right]$.

The general term of the series is $a_k = \left[-\frac{1}{3}-\frac{k}{100}\right]$ for $k = 0, 1, 2, \ldots, 99$. There are $100$ terms in the series.

We need to evaluate the value of $\left[-\frac{1}{3}-\frac{k}{100}\right]$ for each $k$ from $0$ to $99$. The value of $[x]$ is the greatest integer less than or equal to $x$. Let's find the range of $k$ for which the value of $\left[-\frac{1}{3}-\frac{k}{100}\right]$ is constant. We know that $[x] = n$ if $n \le x < n+1$ for an integer $n$. So, $\left[-\frac{1}{3}-\frac{k}{100}\right] = n$ if $n \le -\frac{1}{3}-\frac{k}{100} < n+1$. Multiplying by $-1$ and reversing the inequalities, we get $-n-1 < \frac{1}{3}+\frac{k}{100} \le -n$.

Let's find the values of $n$ that the terms can take. For $k=0$, the term is $\left[-\frac{1}{3}\right] = -1$. So $n=-1$ occurs. For the term to be $-1$, we have $-1 \le -\frac{1}{3}-\frac{k}{100} < 0$. $1 \ge \frac{1}{3}+\frac{k}{100} > 0$. $\frac{2}{3} \ge \frac{k}{100} > -\frac{1}{3}$. $\frac{200}{3} \ge k > -\frac{100}{3}$. $66.66\ldots \ge k > -33.33\ldots$. Since $k$ starts from $0$, the values of $k$ for which $\left[-\frac{1}{3}-\frac{k}{100}\right] = -1$ are $k=0, 1, \ldots, 66$. The number of such terms is $66 - 0 + 1 = 67$. The sum of these terms is $67 \times (-1) = -67$.

Let's find when the value of the term is $-2$. So $n=-2$. For the term to be $-2$, we have $-2 \le -\frac{1}{3}-\frac{k}{100} < -1$. $2 \ge \frac{1}{3}+\frac{k}{100} > 1$. $2-\frac{1}{3} \ge \frac{k}{100} > 1-\frac{1}{3}$. $\frac{5}{3} \ge \frac{k}{100} > \frac{2}{3}$. $\frac{500}{3} \ge k > \frac{200}{3}$. $166.66\ldots \ge k > 66.66\ldots$. The values of $k$ in this range that are in our series ($0 \le k \le 99$) are $k=67, 68, \ldots, 99$. The number of such terms is $99 - 67 + 1 = 33$. The sum of these terms is $33 \times (-2) = -66$.

Let's check if any term is $-3$. So $n=-3$. For the term to be $-3$, we have $-3 \le -\frac{1}{3}-\frac{k}{100} < -2$. $3 \ge \frac{1}{3}+\frac{k}{100} > 2$. $3-\frac{1}{3} \ge \frac{k}{100} > 2-\frac{1}{3}$. $\frac{8}{3} \ge \frac{k}{100} > \frac{5}{3}$. $\frac{800}{3} \ge k > \frac{500}{3}$. $266.66\ldots \ge k > 166.66\ldots$. Since the maximum value of $k$ in the series is $99$, there are no terms with value $-3$ or less.

The total sum of the series is the sum of the terms with value $-1$ and the sum of the terms with value $-2$. Total sum $S = (-67) + (-66) = -133$.