For $$x \in R$$, Let [x] denotes the greatest integer $$\leq x$$, then the sum of the series $$\left[-\frac{1}{3}\right] + \left[-\frac{1}{3} - \frac{1}{100}\right] + \left[-\frac{1}{3} - \frac{2}{100}\right] + \ldots + \left[-\frac{1}{3} - \frac{99}{100}\right]$$ is

We begin by recalling the definition of the greatest-integer (floor) function. By definition, for any real number $$x$$

$$[x]=n \quad\text{if and only if}\quad n\le x<n+1,$$

where $$n$$ is an integer. In words, $$[x]$$ is the largest integer that is not greater than $$x$$.

The series to be summed is

$$\left[-\frac13\right]+\left[-\frac13-\frac1{100}\right]+\left[-\frac13-\frac2{100}\right]+\ldots+\left[-\frac13-\frac{99}{100}\right].$$

To handle all the terms uniformly, let us label the general (k-th) term:

$$T_k=\left[-\frac13-\frac{k}{100}\right], \qquad k=0,1,2,\ldots ,99.$$

Our task is to find $$T_k$$ for every integer $$k$$ in the stated range and then add all the $$T_k$$.

First, convert $$\displaystyle -\frac13$$ into its decimal form for easy comparison:

$$-\frac13=-0.333333\ldots$$

Next, observe that

$$-\frac13-\frac{k}{100}=-0.333333\ldots-\frac{k}{100}.$$

Because $$\dfrac{k}{100}$$ varies from $$0$$ (when $$k=0$$) to $$\dfrac{99}{100}=0.99$$ (when $$k=99$$), the expression

$$\alpha_k=-\frac13-\frac{k}{100}$$

will move from

$$\alpha_0=-0.333333\ldots$$

down to

$$\alpha_{99}=-0.333333\ldots-0.99=-1.323333\ldots$$

Thus each $$\alpha_k$$ lies in the interval

$$-1.323333\ldots\le\alpha_k\le-0.333333\ldots$$

Within this interval the only integers that can serve as greatest integers are $$-1$$ and $$-2$$, because

$$-2<-1.323333\ldots\le\alpha_k\le-0.333333\ldots<0.$$

So every term $$T_k$$ equals either $$-1$$ or $$-2$$. To decide which, locate the point where $$\alpha_k$$ crosses $$-1$$. We need

$$\alpha_k<-1\quad\Longleftrightarrow\quad -\frac13-\frac{k}{100}<-1.$$

Rearranging, add $$\dfrac13$$ to both sides:

$$-\frac{k}{100}<-1+\frac13=-\frac23.$$

Multiplying by $$-1$$ reverses the inequality:

$$\frac{k}{100}>\frac23.$$

Now convert $$\dfrac23$$ to a decimal:

$$\frac23=0.666666\ldots$$

Hence

$$\frac{k}{100}>0.666666\ldots\quad\Longleftrightarrow\quad k>66.6666\ldots$$

Since $$k$$ is an integer, the inequality becomes

$$k\ge 67.$$

We have therefore obtained two cases.

• If $$0\le k\le 66$$, then $$\alpha_k\ge-1$$, so

$$T_k=[\alpha_k]=-1.$$

• If $$67\le k\le 99$$, then $$\alpha_k<-1$$, but always $$\alpha_k>-2$$, so

$$T_k=[\alpha_k]=-2.$$

Count how many times each value occurs.

• For $$k=0,1,2,\ldots,66$$ we have $$67$$ integers, giving $$67$$ occurrences of $$-1$$.

• For $$k=67,68,\ldots,99$$ we have $$99-67+1=33$$ integers, giving $$33$$ occurrences of $$-2$$.

Now add all the terms:

$$\text{Sum}=67(-1)+33(-2).$$

Simplify step by step:

$$67(-1)=-67,$$

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

and finally

$$-67-66=-133.$$

Hence, the correct answer is Option D.