Question

For $x\in R$, let [x] denote greatest integer$\le x$, then the sum of the series $\left[ \dfrac{-1}{3} \right]+\left[ \dfrac{-1}{3}-\dfrac{1}{100} \right]+\left[ \dfrac{-1}{3}-\dfrac{2}{100} \right]+....+\left[ \dfrac{-1}{3}-\dfrac{99}{100} \right]$ is:
(a) -153
(b) -133
(c) -131
(d) -135

Answer 1

Hint: First, we can see that all terms are in the form of the denominator as 100, so we multiply and divide the first term by 100 and also solving the further terms by using LCM. Then, we can see clearly it follows the arithmetic progression where a is -100 and d is -3. Then, by using the formula for the nth term of an AP and substituting ${{a}_{n}}$as 300 to get the two ranges of greatest integer function and then by using various values around it we get the dividing range of the greatest integer function which helps us to calculate the sum of the given series.

Complete step by step answer:
In this question, we are supposed to find the sum of the series $\left[ \dfrac{-1}{3} \right]+\left[ \dfrac{-1}{3}-\dfrac{1}{100} \right]+\left[ \dfrac{-1}{3}-\dfrac{2}{100} \right]+....+\left[ \dfrac{-1}{3}-\dfrac{99}{100} \right]$ for the condition that $x\in R$, let [x] denote greatest integer$\le x$.
So, before proceeding for this, we must know the form of the greatest integer function which can be represented as a summation of the integer part with the decimal part as:
$x=\left[ x \right]+k$
Here, we have [x] belongs to the integer part and k belongs the range as (0So, to understand it more, we can take two examples as:
$\begin{align}
  & 2.13=2+0.13 \\
 & -0.7=-1+0.3 \\
\end{align}$
So, we can see clearly in 2.13, 0.13 is the k part which is in decimal and 2 is the integer part and same rule for -0.7.
Now, applying same concept to the given series, we get:
$\left[ \dfrac{-1}{3} \right]+\left[ \dfrac{-1}{3}-\dfrac{1}{100} \right]+\left[ \dfrac{-1}{3}-\dfrac{2}{100} \right]+....+\left[ \dfrac{-1}{3}-\dfrac{99}{100} \right]$
Here, we can see that all terms are in form of denominator as 100, so we multiply and divide the first term by 100 and also solving the further terms by using LCM, we get:
$\begin{align}
  & \left[ \dfrac{-100}{300} \right]+\left[ \dfrac{-100-3}{300} \right]+\left[ \dfrac{-100-6}{300} \right]+....+\left[ \dfrac{-100-297}{300} \right] \\
 & \Rightarrow \left[ \dfrac{-100}{300} \right]+\left[ \dfrac{-103}{300} \right]+\left[ \dfrac{-106}{300} \right]+....+\left[ \dfrac{-397}{300} \right] \\
\end{align}$
So, by using only the numerator of the series to get the relation for the greatest integer function as:
$\left( -100 \right),\left( -103 \right),\left( -106 \right),.....,\left( -397 \right)$
So, we can see clearly it follows the arithmetic progression where a is -100 and d is -3.
Now, by using the formula for the nth term of an AP and substituting ${{a}_{n}}$as 300 to get the two ranges of greatest integer function as:
$\begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow -300=-100+\left( n-1 \right)\left( -3 \right) \\
 & \Rightarrow -300+100=-3n+3 \\
 & \Rightarrow -200-3=-3n \\
 & \Rightarrow -203=-3n \\
 & \Rightarrow n=\dfrac{203}{3} \\
 & \Rightarrow n=67.66 \\
\end{align}$
So, we calculate the value of 66 term of the AP, we get:
$\begin{align}
  & {{a}_{66}}=-100+\left( 66-1 \right)\left( -3 \right) \\
 & \Rightarrow {{a}_{66}}=-100+\left( 65 \right)\left( -3 \right) \\
 & \Rightarrow {{a}_{66}}=-100-195 \\
 & \Rightarrow {{a}_{66}}=-295 \\
\end{align}$
Similarly, we calculate the value of 67 term of the AP, we get:
$\begin{align}
  & {{a}_{67}}=-100+\left( 67-1 \right)\left( -3 \right) \\
 & \Rightarrow {{a}_{67}}=-100+\left( 66 \right)\left( -3 \right) \\
 & \Rightarrow {{a}_{67}}=-100-198 \\
 & \Rightarrow {{a}_{67}}=-298 \\
\end{align}$
Also, we calculate the value of 68 term of the AP, we get:
$\begin{align}
  & {{a}_{68}}=-100+\left( 68-1 \right)\left( -3 \right) \\
 & \Rightarrow {{a}_{68}}=-100+\left( 67 \right)\left( -3 \right) \\
 & \Rightarrow {{a}_{68}}=-100-201 \\
 & \Rightarrow {{a}_{68}}=-301 \\
\end{align}$
So, we can conclude from the above calculations that the series will the value (-1) for greatest integer function till the 67 terms of the series and after that, we get the value as (-2) for the greatest integer function till $100th$ term.
So, by using the above conclusion, we get the summation of series for total 100 terms as:
$\begin{align}
  & 67\left( -1 \right)+33\left( -2 \right) \\
 & \Rightarrow -67-66 \\
 & \Rightarrow -133 \\
\end{align}$
So, we get the sum of the series $\left[ \dfrac{-1}{3} \right]+\left[ \dfrac{-1}{3}-\dfrac{1}{100} \right]+\left[ \dfrac{-1}{3}-\dfrac{2}{100} \right]+....+\left[ \dfrac{-1}{3}-\dfrac{99}{100} \right]$as -133.
Hence, (b) is correct.

Note:
Now, to solve these type of the questions we need to know some of the basic rules of the greatest integer function to solve the question appropriately. So, the basic rules of greatest integer function are:
If $x\in \left[ -1,0 \right)$then [x]=-1
If $\left[ -2,-1 \right)$then [x]=-2
Also, the formula for the AP with first term a and common difference d for ant nth term ${{a}_{n}}$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$.